MATHS MCQ

$ = \frac{{( - 36)( - 14) - (23)( - 2)}}{{(2)(23) - {{(36)}^2}}},\;\frac{{( - 2)( - 36) - (2)( - 14)}}{{(2)(23) - {{( - 36)}^2}}}$

$ = \left( { - \frac{{11}}{{25}},\, - \frac{2}{{25}}} \right)$.

$i.e.$, $\left( {\frac{{hf - bg}}{{ab - {h^2}}},\,\frac{{gh - af}}{{ab - {h^2}}}} \right)$.

Then $\sqrt {{{(x - 1)}^2} + {{(y + 1)}^2}} = \sqrt 2 \left( {\frac{{x - y + 1}}{{\sqrt 2 }}} \right)$,

[Using $SP = e.PM]$

$ \Rightarrow 2xy - 4x + 4y + 1 = 0$.

Squaring both sides, we get $\sqrt {{{(x + 2)}^2} + {y^2}} = x + 2$

Again squaring both sides, we get ${y^2} = 0$, which is the equation of pair of straight lines.

Comparing the given equation with

$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$

We get, $a = 1$, $h = 0$, $b = 1$, $g = 1$, $f = 0$, $c = - 1$

$\therefore $ $\Delta = abc + 2fgh - a{f^2} - b{g^2} - c{h^2}$

$\Delta = 1 + 0 + 0 - 1 = 0$

Hence, the given equation represents two straight lines.

$ = 2 - \frac{9}{4} - 2 < 0$ and ${h^2} - ab = 1 - 1 = 0$.

i.e., ${h^2} = ab$ ==> a parabola.

Similarly eliminating s from $x = 2s,\,y = \frac{2}{s},$

we get $xy = 4.$

Hence point of intersection is $(2, 2).$

$\frac{{{{(x - 1)}^2}}}{2} + {\left( {y + \frac{3}{4}} \right)^2} = \frac{1}{{16}}$

==> $\frac{{{{(x - 1)}^2}}}{{(1/8)}} + \frac{{{{\left( {y + \frac{3}{4}} \right)}^2}}}{{(1/16)}} = 1$,

which is an ellipse with ${a^2} = \frac{1}{8}$ and ${b^2} = \frac{1}{{16}}$

$\therefore \frac{1}{{16}} = \frac{1}{8}(1 - {e^2})$

==>${e^2} = 1 - \frac{1}{2}$

==> $e = \frac{1}{{\sqrt 2 }}$.

$ \Rightarrow $ ${(2x - 6)^2} + {(4y - 4)^2} = 53$

$ \Rightarrow $ $4\,{(x - 3)^2} + 16\,{(y - 1)^2} = 53$

$ \Rightarrow $ $\frac{{{{(x - 3)}^2}}}{{53/4}} + \frac{{{{(y - 1)}^2}}}{{53/16}} = 1$

$\therefore $ $e = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} = \sqrt {1 - \frac{{53/16}}{{53/4}}}$

$= \sqrt {1 - \frac{1}{4}} = \frac{{\sqrt 3 }}{2}$.

Focus $S( - 2,\,0)$,

Equation of directrix $2x - 9 = 0$

${(PS)^2} = \frac{4}{9}{(PM)^2}$

==> ${(h + 2)^2} + {(k)^2} = \frac{4}{9}{\left( {\frac{{2h - 9}}{2}} \right)^2}$

==> $9[{(h + 2)^2} + {(k)^2}] = \frac{{4{{(2h - 9)}^2}}}{4}$

==> $9{h^2} + 9{k^2} + 36h + 36 = 4{h^2} + 81 + 36h$

==> $\frac{{5{h^2}}}{{45}} + \frac{{9{k^2}}}{{45}} = 1$

==> $\frac{{{h^2}}}{9} + \frac{{{k^2}}}{5} = 1$ 

Locus of point $P(h, k)$ is $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{5} = 1$, which is an ellipse

$i.e.$, a hyperbola.

$\Delta = (1)\,(1)\,(2) + 2(2)\,(1)\,(2) - (1)\,{(2)^2} - (1)\,{(1)^2} - 2{(2)^2} < 0$

Hence it is a hyperbola.

$⇒$  $ - \frac{{{{(x + 1)}^2}}}{4} + \frac{{{y^2}}}{4} = 1$

Equation of directrices of $\frac{{{y^2}}}{{{b^2}}} - \frac{{{x^2}}}{{{a^2}}} = 1$ are $y = \pm \frac{b}{e}$

Here $b = 2,\;e = \sqrt {1 + 1} = \sqrt 2 $

Hence $y = \pm \frac{2}{{\sqrt 2 }}$

$⇒$ $y = \pm \sqrt 2 $.

Now let the various point be $(h,k)$,

then accordingly $\frac{{\sqrt {{{(h - 1)}^2} + {{(k - 1)}^2}} }}{{\frac{{2h + k - 1}}{{\sqrt 5 }}}} = \sqrt 3 $

Squaring both the sides, we get

$5[{(h - 1)^2} + {(k - 1)^2}] = 3{(2h + k - 1)^2}$

On simplification, the required locus is

$7{x^2} + 12xy - 2{y^2} - 2x + 4y - 7 = 0$.

==> $({x^2} - 2x) - 4({y^2} - 4y) - 40 = 0$

==> ${(x - 1)^2} - 1 - 4[{(y - 2)^2} - 4] - 40 = 0$

==> ${(x - 1)^2} - 4{(y - 2)^2} = 25$

==> $\frac{{{{(x - 1)}^2}}}{{25}} - \frac{{{{(y - 2)}^2}}}{{25/4}} = 1$, which is a hyperbola.

$h = \frac{{a\,\cos \alpha - b\,\sin \alpha }}{{\cos 2\alpha }}$, $k = \,\frac{{a\,\sin \alpha - \,b\cos \alpha }}{{\cos 2\alpha }}$

Then the locus of point $(h,\,k)$ is ${x^4} + {y^4} - 2{x^2}{y^2}$ $ = ({a^2} + {b^2})$ $({x^2} + {y^2}) + 4abxy$, which is not a locus of any given curves.

Obviously, it is a parabola whose focus is $\left( {\frac{5}{2},1} \right)$ and directrix

is $x = \frac{3}{2}$.

==> $9{y^2} = 4ax$.

Hence vertex is $(1, -1)$, which lies in $IV$ quadrant.

Hence the vertex is $\left( {\frac{1}{3},\, - \frac{2}{9}} \right)$.

Passes through $(3, 6)$ 

==> $16 = 4a.8$

==> $a = \frac{1}{2}$

==> ${(x + 1)^2} = 2(y + 2)$

==> ${x^2} + 2x - 2y - 3 = 0$.

$\frac{{\sqrt {{{(x + 8)}^2} + {{(y + 2)}^2}} }}{{\left| {\frac{{2x - y - 9}}{{\sqrt 5 }}} \right|}} = 1$

Squaring and after simplification, we get

${x^2} + 4{y^2} + 4xy + 116x + 2y + 259 = 0$.

Focus is $(-3, 0)$ 

$⇒$  $2a = ( - 5 + 3) = 2$ $⇒$  $a = 1$

Vertex is $\left( {\frac{{ - 3 + ( - 5)}}{2},\,0} \right) = ( - 4,\,0)$

Therefore, equation is ${(y - 0)^2} = 4(x + 4)$.

where $A = (a' - a)$ or ${y^2} = 4(a' - a)(x - a)$.

==> ${(y - 2)^2} = - 4(x - 1)$

Vertex is $(1,2)$ and focus = $(0,2)$.

==> ${(x + 2)^2} = - 2\left( {y - \frac{{11}}{2}} \right)$

Hence vertex is $\left( { - 2,\frac{{11}}{2}} \right)$.

So none is correct.

Equation of latus rectum $x + \lambda = 0$

Vertex $ = \left( {\frac{{ - C + {B^2}}}{{2A}},B} \right)$, focus $ \equiv \left( {\frac{{ - C + {B^2}}}{{2A}} - \frac{A}{2},B} \right)$

Equation of $L.R.$ is $x = \frac{{ - C + {B^2}}}{{2A}} - \frac{A}{2} = \frac{{{B^2} - {A^2} - C}}{{2A}}$.

$\therefore $ $a = 2$

Hence parabola is ${(x - 0)^2} = - 4.2(y - 4)$

$i.e.$,${x^2} + 8y = 32$.

==> ${(3x - 1)^2} = - 36y - 18 = - 36\left( {y + \frac{1}{2}} \right)$

==> $9{\left( {x - \frac{1}{3}} \right)^2} = - 36\left( {y + \frac{1}{2}} \right)$

Hence length of latus rectum is $4$.

==> ${\left( {y - \frac{2}{3}} \right)^2} = \frac{{16}}{9}\left( {x + \frac{{61}}{{16}}} \right)$

Put $y - \frac{2}{3} = Y$and $x + \frac{{61}}{{16}} = X$

==> ${Y^2} = 4\left( {\frac{4}{9}} \right)\,X$

Axis of this parabola is $Y = 0$

==> $y - \frac{2}{3} = 0$

==> $3y = 2$.

where $x = X + a,\,\,y = Y + b$.

Therefore the equation of the parabola referred to the point $(a,b)$ as the vertex is

${(x - a)^2} = l(y - b)$ or ${(x - a)^2} = \frac{l}{2}(2y - 2b)$.

${(y - 2)^2} = 5(x + 1)$

Obviously, latus rectum is $5$.

or ${x^2} - 2x + 1 = 2y + 1$

==> ${(x - 1)^2} = 2\left( {y + \frac{1}{2}} \right)$

Here $4a = 2 $

$\Rightarrow a = \frac{1}{2}$

Now focus is $\left( {x - 1 = 0,\,y + \frac{1}{2} = \frac{1}{2}} \right)\,\,i.e.,\,(1,\,0)$.

${y^2} - 4y - 2x - 8 = 0$

==> ${(y - 2)^2} = 2(x + 6)$

$\therefore $Length of latus rectum = $2$.

On solving we get

${(ax - by)^2} - 2{a^3}x - 2{b^3}y + {a^4} + {a^2}{b^2} + {b^4} = 0$.

$2{\left( {y - \frac{5}{2}} \right)^2} = - (x - 4)$

==> $4a = \frac{1}{2}$.

${(y + 1)^2} = \frac{3}{2}(x + 3).$

So, vertex is $( - 3, - 1)$.

$4{\left( {y - \frac{1}{2}} \right)^2} = 6(x + 1)$

$\Rightarrow {\left( {y - \frac{1}{2}} \right)^2} = \frac{3}{2}(x + 1) $

$\Rightarrow {Y^2} = \frac{3}{2}X$

where, $Y = y - \frac{1}{2},\,\,X = x + 1$

$\therefore y = Y + \frac{1}{2},\,\,\,x = X - 1$…..$(i)$

For focus $X = a,\,\,Y = 0$

$\because 4a = \frac{3}{2} \Rightarrow a = \frac{3}{8} \Rightarrow x = \frac{3}{8} - 1 =  - \frac{5}{8}$

$y = 0 + \frac{1}{2} = \frac{1}{2}$, Focus$ = \left( { - \frac{5}{8},\frac{1}{2}} \right)$.

It can be written as ${(x + 4)^2} = - 12y + 12$

$ \Rightarrow \,{(x + 4)^2} = - 12(y - 1)$,  vertex is $( - \,4,\,1).$

Comparing the given equation with the standard equation, we get vertex $(\alpha ,\,\beta ) = ( - 3,\,2)$ and $a = 5.$

Therefore focus of the parabola $(\alpha + a,\,\beta ) = (2,\,2)$.

$ \Rightarrow $${x^2} - 4x = 8y - 12$

$ \Rightarrow $${(x - 2)^2} = 8(y - 1)$

Hence the length of latus rectum =$4a = 8$.

$y = 2{x^2} + x$

$ \Rightarrow \,{x^2} + \frac{x}{2} = \frac{y}{2}$

$ \Rightarrow \,{\left( {x + \frac{1}{4}} \right)^2} = \frac{y}{2} + \frac{1}{{16}}$

$ \Rightarrow \,{\left( {x + \frac{1}{4}} \right)^2} = \frac{1}{2}\left( {y + \frac{1}{8}} \right)$

It can be written as, ${X^2} = \frac{1}{2}Y$.....$(i)$

Here $A = \frac{1}{8}$, focus of $(i)$ is $\left( {0,\frac{1}{8}} \right)$

$i.e.$ $X = 0$, $Y = \frac{1}{8}$

==> $x + \frac{1}{4} = 0$, $y + \frac{1}{8} = \frac{1}{8}$

$ \Rightarrow \,x = - \frac{1}{4},$ $y = 0$

$i.e.$ focus of given parabola is $\left( { - \frac{1}{4},\,0} \right)$.

${y^2} - 2y - x + 2 = 0$

${(y - 1)^2} = (x - 1)$

Let $y - 1 = Y$ and $x - 1 = X$

${Y^2} = X,\,a = 1/4$, focus $ = (1/4,\,0)$

Required focus = $\left( {\frac{1}{4} + 1,\,0 + 1} \right) = (5/4,\,1)$.

Comparing the given equation with the standard equation, we get $h = 1,\,k = 2$ and $4a = 16$ or $a = 4.$

Therefore vertex of the parabola $ \equiv \,(h,\,k) \equiv (1,\,2)$.

We know that as the $y$ - coordinates of vertex and focus are same,

therefore axis of parabola is parallel to $x$ - axis.

Thus equation of the parabola is

${(y - \,k)^2} = 4a\,(x - h)$ or ${(y - 1)^2}$ $ = 4 \times 2(x - 1)$ or ${(y - 1)^2} = 8(x - 1).$

$ \Rightarrow $ ${(x - 5)^2} + {(y - 3)^2} = {\left( {\frac{{3x - 4y + 1}}{{\sqrt {9 + 16} }}} \right)^2}$

==> $25({x^2} + 25 - 10x + {y^2} + 9 - 6x)$

$ = 9{x^2} + 16{y^2} + 1 - 12xy + 6x - 8y - 12xy$

==> $16{x^2} + 9{y^2} - 256x - 142y + 24xy + 849 = 0$

$ \Rightarrow $ ${(4x + 3y)^2} - 256x - 142y + 849 = 0.$

From ${(x - h)^2} = - 4a(y - k)$

Parabola is,

${(x - 2)^2} = - 4.2(y + 1)$

==> ${(x - 2)^2} = - \,8(y + 1)$

==> ${x^2} + 4 - 4x = - 8y - 8$

==> ${x^2} - 4x + 8y + 12 = 0.$

==> ${x^2} - 4x + 4 = 8y - 8$

==> ${(x - 2)^2} = 8(y - 1)$

==> ${X^2} = 8Y$

Comparing with ${X^2} = 4aY$, we get $a = 2$

Directrix is $Y = - a$

==> $y - 1 = - 2$

==> $y = - 1$.

==> ${x^2} + {y^2} = {\left( {\frac{{x + y - 4}}{{\sqrt 2 }}} \right)^2}$

==> ${x^2} + {y^2} - 2xy + 8x + 8y - 16 = 0.$

then $Z \equiv (0,\,9)$ $i.e.$, $y = 9$

$\therefore $ Equation of parabola, $SP = PM$

$ \Rightarrow $ $\sqrt {{{(x - 0)}^2} + {{(y - 3)}^2}} = |y - 9|$

$ \Rightarrow $ ${x^2} + {y^2} - 6y + 9 = {y^2} - 18y + 81$

or ${x^2} + 12y = 72$.

$ \Rightarrow $${x^2} - 2x + 1 + 8y - 7 - 1 = 0$ $ \Rightarrow $ ${(x - 1)^2} + 8y = 8$

$ \Rightarrow $ ${(x - 1)^2} = - 8(y - 1)$ $ \Rightarrow $${(x - 1)^2} = - 4.2(y - 1)$

Here, $a = 2$.

$\therefore $ Equation of directrix is $y - 1 = 2 $ $i.e.$, $y = 3$.

$ \Rightarrow $ ${x^2} - \frac{3}{2}x = - \frac{5}{2}y - 2$

$ \Rightarrow $ ${\left( {x - \frac{3}{4}} \right)^2} = - \frac{5}{2}y - \frac{{23}}{{16}}$

$\therefore $ Equation of axis is, $x - \frac{3}{4} = 0$ 

$ \Rightarrow $ $x = \frac{3}{4}$.

$ \Rightarrow $${x^2} + 6x = - 20y + 51$

$ \Rightarrow $${(x + 3)^2} = - 20y + 60$

$\Rightarrow {(x + 3)^2} = - 20(y - 3)$

$ \Rightarrow $${(x + 3)^2} = - 4.5(y - 3)$

$\therefore $ Axis of parabola is $x + 3 = 0$.

==> ${\left( {\frac{{dy}}{{dx}}} \right)_{(1,\,0)}} = 2 - 1 = 1$

Hence the tangent is $y - 0 = x - 1$.

Therefore ${(y + 1)^2} = - (4x - 9) = - 4\left( {x - \frac{9}{4}} \right)$

$S \equiv \left( { - 1 + \frac{9}{4},\, - 1} \right)$ or $\left( {\frac{5}{4},\, - 1} \right)$.

then tangent at this point is $ky = x + h$.

$x - ky + h = 0 \equiv 18x - 6y + 1 = 0$

or $\frac{1}{{18}} = \frac{k}{6} = \frac{h}{1}$

or $k = \frac{1}{3}$, $h = \frac{1}{{18}}$.

Tangent parallel to $Y = 2X + 7$ is $Y = 2X + \frac{a}{m}$

==> $y = 2\left( {x + \frac{5}{4}} \right) + \frac{1}{2}$

==> $y = 2x + 3$ $i.e.$, $2x - y + 3 = 0$.

$8y = 2(x + 16)$.....$(i)$ and $16x = 16(y + 8)$.....$(ii)$

 ${m_1} = \frac{1}{4},\,\,{m_2} = 1$

$\tan \theta = \frac{{{m_2} - {m_1}}}{{1 + {m_2}{m_1}}} = \left( {\frac{3}{5}} \right)$

==> $\theta = {\tan ^{ - 1}}\left( {\frac{3}{5}} \right)$.

Aliter : Using direct formula

$\theta = {\tan ^{ - 1}}\frac{{3{a^{1/3}}{b^{1/3}}}}{{2({a^{2/3}} + {b^{2/3}})}},$

where $a = 1$ and $b = 8$

$ = {\tan ^{ - 1}}\frac{6}{{2(1 + 4)}} = {\tan ^{ - 1}}\frac{3}{5}$.

Also from equation of parabola, we get gradient at $(h,k)$ as the slope of parabola

$ = \frac{{dy}}{{dx}} = \frac{{ - 1}}{{2y - 1}} = \frac{{ - 1}}{{2k - 1}}$

Since line and parabola touch at $(h,k)$

==> $\frac{{ - 1}}{{2k - 1}} = - 1$

==> $ - 2k + 1 = - 1$

==> $k = 1$

Putting this value in $x + y = 1$, we have $h = 0,$

so the point of contact is $(0,\,1).$

Therefore, tangent at ${y^2} = 4a(x + a)$ is,

$y = m(x + a) + \frac{a}{m}$

or $y = mx + ma + \frac{a}{m}$

==> $ma + \frac{a}{m} = c$.

Slope of tangent $ = \frac{{3 \pm 1}}{{1 \mp 3}} = - 2$ or $\frac{1}{2}$

Tangent is $y = - 2x + \frac{2}{{ - 2}}$ or $2x + y + 1 = 0$.

Since it is a tangent to ${y^2} = 4a(x - a),$ it will cut it in two coincident points.

 Roots of ${m^2}{x^2} - 4ax + 4{a^2} = 0$ are equal.

$16{a^2} - 16{a^2}{m^2} = 0$ or ${m^2} = 1$ or $m = 1, - 1$

Product of slopes $ = - 1$.

Hence it is a right angled triangle.

Therefore the co-ordinates of $P$ are $(4,4)$.

The equations of the tangents to the two parabolas at $(4,4)$ are $2x - y - 4 = 0$.....$(i)$

$x - 2y + 4 = 0$.....$(ii)$

Now, ${m_1} = $Slope of $(i)$ $ = 2,$${m_2} = $Slope of $(ii)$ $ = \frac{1}{2}$

${m_1}{m_2} = 1\,\,\,i.e.,\,\,\,\tan {\theta _1}\tan {\theta _2} = 1$.

$ty = x + a{t^2}$ .....$(i)$

Clearly, $lx + my + n = 0$ is also a chord of contact of tangents.

Therefore $ty = x + a{t^2}$ and $lx + my + n = 0$ represents the same line.

Hence, $\frac{1}{l} = - \frac{t}{m} = \frac{{a{t^2}}}{n}$

==> $t = \frac{{ - m}}{l},\,\,{t^2} = \frac{n}{{la}}$

Eliminating $t$, we get, ${m^2} = \frac{{nl}}{a}$

$i.e.$, an equation of parabola.

Since tangents are perpendicular to the parabola,

therefore, $\frac{1}{{{t_1}}}.\frac{1}{{{t_2}}} = - 1$ or ${t_1}{t_2} = - 1$.

We also know that their point of intersection $ = (a{t_1}{t_2},\,a({t_1} + {t_2}))$ $ = \,( - a,\,a({t_1} + {t_2})).$

Thus these points lie on directrix $x = - \,a$ or $x + a = 0$.

If the tangent at $P$, $ty = x + a{t^2},$ meets the directrix

$x = - a\,\,{\rm{at}}\,\,k,$ then $k = \left( { - a,\,\frac{{a{t^2} - a}}{t}} \right)$

${m_1} = $ slope of $SP = \frac{{2at}}{{a({t^2} - 1)}}$

${m_2} = $ slope of $SK = \frac{{a({t^2} - 1)}}{{ - 2at}}$

Clearly ${m_1}{m_2} = - 1$,

$\therefore \,\angle \,PSK = {90^o}.$

$\frac{{dy}}{{dx}} = \frac{{2x}}{4}$ and $\frac{{ - 2x}}{4}$

${m_1} = 0,\,\,{m_2} = 0$

==> $\theta = {0^o}.$

therefore angle between them is ${90^o}$.

$i.e.$, ${y^2} = 4\left( {\frac{a}{4}} \right)\,x$.....(i)

Let point of contact is $({x_1},\,{y_1})$

Equation of tangent is $y - {y_1} = \frac{{2\left( {\frac{a}{4}} \right)}}{{{y_1}}}\,(x - {x_1})$

$⇒$  $y = \frac{a}{{2{y_1}}}(x)\, - \frac{{a{x_1}}}{{2{y_1}}} + {y_1}$

Here, $m = \frac{a}{{2{y_1}}} = \,\tan \,{45^o}$$ \Rightarrow \,$$\frac{a}{{2{y_1}}} = 1$

$⇒$  ${y_1} = \frac{a}{2}$

From (i), ${x_1} = \frac{a}{4}$. 

Point is $\left( {\frac{a}{4},\,\frac{a}{2}} \right)$.

It intersects ${y^2} = 4ax$ in $\left( {\frac{{4a}}{{{m^2}}},\frac{{4a}}{m}} \right)$.

Mid point of the chord is $\left( {\frac{{2a}}{{{m^2}}},\frac{{2a}}{m}} \right)$

$x = \frac{{2a}}{{{m^2}}},$$y = \frac{{2a}}{m}$

==> $\frac{{2a}}{x} = \frac{{4{a^2}}}{{{y^2}}}$ or ${y^2} = 2ax$, which is a parabola.

Normal is $y - k = \frac{{ - k}}{4}(x - h)$ or $ - kx - 4y + kh + 4k = 0$

Gradient $ = - \frac{k}{4} = \frac{1}{2}$

==> $k = - 2$

Substituting $(h,k)$ and $k = - 2$, we get $h = \frac{1}{2}$

Hence point is $\left( {\frac{1}{2}, - 2} \right)$.

Trick : Here only point $\left( {\frac{1}{2}, - 2} \right)$ satisfies the parabola ${y^2} = 8x$.

$y - k = - \frac{k}{4}(x - h)$

Gradient$ = \tan 60^\circ = \sqrt 3 = - \frac{k}{4}$

==> $k = - 4\sqrt 3 $ and $h = 6$

Hence required point is $(6, - 4\sqrt 3 )$.

 => $2y + x = 2a + \frac{a}{4} = \frac{{9a}}{4}$

=> $2y + x - \frac{{9a}}{4} = 0$

$i.e.$, $4x + 8y - 9a = 0$.

or $ - k = - \{ - 4( - 2) - 2{( - 2)^3}\} = - (8 + 16)$

==> $k = 24$

Here $a = \frac{6}{4}$

Therefore, $ - ve$end of latus rectum is $\left( {\frac{3}{2},\, - 3} \right)$

Line through the point is $y = \frac{{ - 3}}{{3/2}}x$or $y + 2x = 0$.

 $b = \frac{{2ac}}{{a + c}}$

==> $a, b, c$ are in $H.P.$

${y^2} = 4ax\left( {\frac{{lx + my}}{{ - n}}} \right)$ or $4al{x^2} + 4amxy + n{y^2} = 0$

This represents a pair of perpendicular lines, if $4al + n = 0$.

If the normal makes equal angles with the coordinates axes, then $m = \tan \frac{\pi }{4} = 1.$

Thus, the required point is $(1, -2).$

$i.e.$, $x = - a$ or $x + a = 0$.

For three values of $m$.

Three normal can be drawn on parabola ${y^2} = 4ax$.

So three feet of normals can be obtained, hence centroid of triangle lies on axis of parabola.

Since $a = 2,\,\,{t_1} = 1$;

$\therefore $ ${t_2} = - 3$

$\therefore $The other end will be $(at_2^2,\,2a{t_2})$ $i.e.$, $(18, -12).$

$a = \frac{1}{4},$$b = 1,\,\,c = 4$. Area of triangle formed by

$\left( {\frac{1}{4},\,1} \right)\,,\,\,(1,\,2),\,\,(4,\,4)$ is $\frac{1}{2}\left| {\begin{array}{*{20}{c}}{1/4}&1&1\\1&2&1\\4&4&1\end{array}} \right| = \frac{3}{4}$.

Use area formula and get $\Delta = \frac{1}{{8a}}({y_1} - {y_2})({y_2} - {y_3})({y_3} - {y_1})$.

$P{Q^2} = 32 + 32 = 64$

==> $PQ = 8$

Also, if $p$ be perpendicular from $(-1,2)$ on $PQ$, then area of triangle is $\frac{1}{2}PQ.\,\,p $

$= \frac{1}{2}.8.\left( {\frac{4}{{\sqrt 2 }}} \right) = 8\sqrt 2 $.

Thus the tangents are at right angle.

It is clear that angle between the curve

= angle between the $x$ - axis and $y$ - axis $ = \pi /2$.

${x^2} + {y^2} + 4y = 0$ …..$(ii)$

Solving $(i)$ and $(ii),$ $x = 0$, $y = 0$;

$x = 2$, $y = - 2$

which means parabola pass through $(0, 0)$ and $(2, -2)$ these points satisfy the parabola ${y^2} = 2x$.

$\frac{{dy}}{{dx}} = a = 1$ given

Also $y = ax + b$ passing through $(0, 1)$, $b = 1$

So required line be $y = x + 1$

Now point of intersection of the line and parabola gives ${x^2} + 2x + 1 = 4x$

==> ${x^2} - 2x + 1$

==> $x = 1$ ,  $y = 2$

Line touch that parabola.

Length intercepted is equal to $0$.

= Slope of the line which passes through $(-1, 0).$

Required equation, $y - 0 = - \sqrt 3 (x + 1)$

$y + \sqrt 3 (x + 1) = 0$.

$2(SM) = 2 \times \frac{{{u^2}}}{{2g}}(1 + \cos 2\alpha ) = \frac{{2{u^2}{{\cos }^2}\alpha }}{g}$.

Therefore, $L.R. = 2$ (Length of the perpendicular from $(3,\,3)$ on

$3x - 4y - 2 = 0)$ i.e., $\left| {\frac{{9 - 12 - 2}}{{\sqrt {9 + 16} }}} \right|$= 2.

and $ - 6y = 6(x + 3)$ ==> $x - y + 3 = 0$ and $x + y + 3 = 0$.

The intersection of these tangents are $x = - 6$, which is the equation of directrix.

==> $y - 2au = \frac{{2av - 2au}}{{a{v^2} - a{u^2}}}(x - a{u^2})$

==> $y - 2au = \frac{{2a(v - u)}}{{a(v - u)(v + u)}}(x - a{u^2})$

==> $y - 2au = \frac{2}{{(v + u)}}(x - a{u^2})$

If this is focal chord, so it would passes through focus $(a,0)$.

==>$0 - 2au = \frac{2}{{v + u}}(a - a{u^2})$ ==>$ - uv - {u^2} = 1 - {u^2}$, $\therefore $ $uv+1=0.$

$\therefore $ Equation of normal at point P is $y = mx - 2am - a{m^3}$

This normal cuts $x$ - axis at $N$ Putting $y = 0$; we get,

$0 = mx - 2am - a{m^3}$

or $mx = am + a{m^3}$

or $x = \frac{{m(2a + a{m^2})}}{m}$

or $x = 2a + a{m^2}$

So, ON = $2a + a{m^2}$ and $OM = a{m^2}$

Length of subnormal = MN

$\therefore $ MN = ON -OM $ = 2a + a{m^2} - a{m^2} = 2a$.

parabola $\mathrm{y}^{2}=4 \mathrm{ax},$ then

$-\frac{c}{b}=\frac{a}{-\frac{a}{b}} \Rightarrow c=b^{2}$

Since it passes through $(-3, 1)$ and $(2, -2)$,

so $\frac{9}{{{a^2}}} + \frac{1}{{{b^2}}} = 1$ and $\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{4}$

==>${a^2} = \frac{{32}}{3}$, ${b^2} = \frac{{32}}{5}$

Hence required equation of ellipse is $3{x^2} + 5{y^2} = 32$.

Trick : Since only equation $3{x^2} + 5{y^2} = 32$ passes through $(-3, 1)$ and $(2, -2)$. Hence the result.

${e^2} = 1 - \frac{{{b^2}}}{{{a^2}}} = \frac{5}{9}$

==> $e = \frac{{\sqrt 5 }}{3}$

Distance between the pins $ = 2ae = 2\sqrt 5\ cm$

Length of string $ = 2a + 2ae = 6 + 2\sqrt 5\ cm$.

$\left( {\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - 1} \right)\,\left( {\frac{{{h^2}}}{{{a^2}}} + \frac{{{k^2}}}{{{b^2}}} - 1} \right) = {\left( {\frac{{hx}}{{{a^2}}} + \frac{{yk}}{{{b^2}}} - 1} \right)^2}$

Pair of tangents will be perpendicular, if

coefficient of ${x^2}$ + coefficient of ${y^2} = 0$

==>$\frac{{{k^2}}}{{{a^2}{b^2}}} + \frac{{{h^2}}}{{{a^2}{b^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}$

==> ${h^2} + {k^2} = {a^2} + {b^2}$

Replace $(h,k)$ by $(x,y)$

==> ${x^2} + {y^2} = {a^2} + {b^2}$.

In the second case, $e'\, = \sqrt {1 - ({b^2}/{a^2})} $

According to the given condition,

$\sqrt {1 - {b^2}/{a^2}} = \sqrt {1 - (25/169)} $

$ \Rightarrow \,b/a = 5/13$,   $(\because \,\,a > 0,\,b > 0)$

$⇒ a/b = 13/5.$

Hence the centre is $(2, 3).$

Therefore, latus rectum$ = \frac{{2{b^2}}}{a} = \frac{{2.4}}{3} = \frac{8}{3}$.

==> ${(2x - 2)^2} + {(y + 1)^2} = - 1 + 4 + 1$

==> $\frac{{{{(x - 1)}^2}}}{1} + \frac{{{{(y + 1)}^2}}}{4} = 1$

==> $e = \sqrt {1 - \frac{1}{4}} $

==> $e = \frac{{\sqrt 3 }}{2}$.

$e = \frac{1}{2}$ $⇒$ $b = 3\sqrt {1 - \frac{1}{4}} = \frac{{3\sqrt 3 }}{2}$.

Also centre is $(7, 0)$

Equation is $\frac{{{{(x - 7)}^2}}}{9} + \frac{{{y^2}}}{{(27/4)}} = 1$

$3{x^2} + 4{y^2} - 42x + 120 = 0$.

Vertex $ = (4, - 3)$ $⇒$  $a = 4 - 2 = 2$

$⇒$ $e = \frac{1}{2}$

$⇒$ $b = a\sqrt {\left( {1 - \frac{1}{4}} \right)} = \frac{2}{2}\sqrt 3 = \sqrt 3 $

Therefore, equation of ellipse with centre $(2, - 3)$ is

$\frac{{{{(x - 2)}^2}}}{4} + \frac{{{{(y + 3)}^2}}}{3} = 1$.

Squaring and simplifying, we get

$7{x^2} + 2xy + 7{y^2} + 10x - 10y + 7 = 0$.

==> ${a^2} = 5,$ ${b^2} = 9$.

Therefore $e = \frac{2}{3}$.

$ \Rightarrow \frac{{{{(x + 1)}^2}}}{9} + \frac{{{{(y + 2)}^2}}}{4} = 1$;

$e = \sqrt {1 - \frac{4}{9}} = \frac{{\sqrt 5 }}{3}$.

==> $\frac{{{{(x - 2)}^2}}}{4} + \frac{{{{(y - 1)}^2}}}{3} = 1$

==> $\frac{{{X^2}}}{4} + \frac{{{Y^2}}}{3} = 1$

 $e = \sqrt {1 - \frac{3}{4}} = \frac{1}{2}$. 

Foci are $\left( {X = \pm 2 \times \frac{1}{2},\,Y = 0} \right)$

$i.e.$, $(x - 2 = \pm 1,\,\,y - 1 = 0)$$ = (3,\,1)$ and $(1,\,1)$.

$25{x^2} + 9{y^2} - 150x - 90y + 225 = 0$

$ \Rightarrow $$25\,{(x - 3)^2} + 9{(y - 5)^2} = 225$

$ \Rightarrow $$\frac{{{{(x - 3)}^2}}}{9} + \frac{{{{(y - 5)}^2}}}{{25}}$ = 1. Here $b > a$

$\therefore $ Eccentricity $e = \sqrt {1 - \frac{{{a^2}}}{{{b^2}}}} = \sqrt {1 - \frac{9}{{25}}} $

$ = \sqrt {\frac{{16}}{{25}}} = \frac{4}{5}$.

$9 = 25(1 - {e^2})$

==> $\frac{9}{{25}} = 1 - {e^2}$

==> ${e^2} = \frac{{16}}{{25}}$

==> $e = \frac{4}{5}$.

$9{x^2} + 4{y^2} - 6x + 4y + 1 = 0$

==>${(3x - 1)^2} + {(2y + 1)^2} = 1$

==> $\frac{{{{\left( {x - \frac{1}{3}} \right)}^2}}}{{\frac{1}{9}}} + \frac{{{{(y + 1)}^2}}}{{\frac{1}{2}}} = 1$.

Here $a = \frac{1}{3}$, $b = \frac{1}{2}$;  $2a = \frac{2}{3}$, $2b = 1.$

Length of axes are $\left( {1,\,\frac{2}{3}} \right)$.

Therefore, equation of tangent is $y = \sqrt 3 x \pm \sqrt {1 + 3 \cdot 16} $

==> $y = \sqrt 3 x \pm 7$.

But it is tangent to the given ellipse, therefore $m = 0,\, - 1$.

Hence tangents are $y = 3$ and $x + y = 5$.

Given ellipse is $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$,

Locus is ${x^2} + {y^2} = 13.$

==> $9{x^2} + 5({y^2} - 6y + 9) = 45$

==> $\frac{{{x^2}}}{5} + \frac{{{{(y - 3)}^2}}}{9} = 1$

$\because {a^2} < {b^2},$ so axis of ellipse on $y$ - axis.

At $y$ axis, put $x = 0$, so we can obtained vertex.

Then $0 + 5{y^2} - 30y = 0$

$y = 0,\,\,y = 6$

Therefore, tangents of vertex $y = 0,\,\,\,y = 6$.

==> $\frac{{(x - {x_1}){a^2}}}{{{x_1}}} = \frac{{(y - {y_1}){b^2}}}{{{y_1}}}$;

$({x_1},{y_1}) \equiv (0,3),\,\,{a^2} = 5,\,\,{b^2} = 9$

$ \Rightarrow \frac{{(x - 0)}}{0}\,\,\,5 = \frac{{(y - 3).9}}{3}$ or $x = 0$ $i.e.$, $y$ - axis.

which is the standard equation of normal at point $({x_1},\,{y_1})$.

In the given ellipse, ${a^2} = 20,\,{b^2} = \frac{{180}}{{16}}$.

Hence the equation of normal at the point $(2,\,3)$ is

$\frac{{x - 2}}{{2/20}} = \frac{{y - 3}}{{48/180}}$ ==> $40\,(x - 2) = 15(y - 3)$

==> $8x - 3y = 7$

==> $3y - 8x + 7 = 0$.

Tangent at point $(3,-2)$ is $\frac{{(3)x}}{9} + \frac{{( - 2)y}}{4} = 1$ or $\frac{x}{3} - \frac{y}{2} = 1$

$\therefore $Normal is $\frac{x}{2} + \frac{y}{3} = k$ and it passes through point $(3,-2)$

$\therefore $$\frac{3}{2} - \frac{2}{3} = k $

$\Rightarrow k = \frac{5}{6}$

$\therefore $Normal is, $\frac{x}{2} + \frac{y}{3} = \frac{5}{6}$.

$ax\sin \theta - by{\rm{cosec }}\theta = {a^2} - {b^2}$.....$(i)$

Comparing equation $(i)$ with $2x - \frac{8}{3}\lambda y = - 3$. We get,

$a\sin \theta = 2$, $b{\rm{ cosec}}\theta = \frac{8}{3}\lambda $ or $ab = \frac{{16}}{3}\lambda $.....$(ii)$

$\because \,a = 1,\,b = 2$; $2 = \frac{{16}}{3}\lambda $ or $\lambda = 3/8$

==> $a = \pm \sqrt 5 \left( {\frac{3}{{\sqrt 5 }}} \right) = \pm 3$

==> ${a^2} = 9$

${b^2} = {a^2}(1 - {e^2}) = 9\left( {1 - \frac{5}{9}} \right) = 4$

Hence, equation of ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$

==> $4{x^2} + 9{y^2} = 36$.
 

$\frac{{{{(x - 3)}^2}}}{{16}} + \frac{{{y^2}}}{{25}} = 1$

$e = \sqrt {1 - \frac{{16}}{{25}}} = \frac{3}{5}$

$\frac{{(x - {x_1}){a^2}}}{{{x_1}}} = \frac{{(y - {y_1}){b^2}}}{{{y_1}}}$

At $G,\,\,y = 0$

==> $x = CG = \frac{{{x_1}({a^2} - {b^2})}}{{{a^2}}}$

At $g,\,\,\,x = 0$

==> $y = Cg = \frac{{{y_1}({b^2} - {a^2})}}{{{b^2}}}$

$\frac{{x_1^2}}{{{a^2}}} + \frac{{y_1^2}}{{{b^2}}} = 1$

==> ${a^2}{(CG)^2} + {b^2}{(Cg)^2} = {({a^2} - {b^2})^2}.$

$\frac{2 b^{2}}{a}=4$      .....$(2)$

$b^{2}=a^{2}\left(1-e^{2}\right)$      .....$(3)$

From $(1),(2)$ and $( 3)$

$a=4, b=2 \sqrt{2}$

So, $a^{2}=16, b^{2}=8$

Equation of ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{8}=1$

$\Rightarrow x^{2}+2 y^{2}=16$

==> ${e^2} = \frac{{{a^2} + {b^2}}}{{{a^2}}}$

==> ${e_1} = \sqrt {1 + \frac{{{a^2}}}{{{b^2}}}} $

==> $e_1^2 = \frac{{{b^2} + {a^2}}}{{{b^2}}}$

==> $\frac{1}{{e_1^2}} + \frac{1}{{{e^2}}} = 1$.

==> $a = 5$, $b = 2\sqrt 5 $.

Hence the required equation of hyperbola is $\frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{20}} = 1$

==> $4{x^2} - 5{y^2} = 100$.

==> $a = 3$ and $\frac{{18}}{{{a^2}}} - \frac{4}{{{b^2}}} = 1$

==> ${b^2} = 4$

Therefore, $e = \sqrt {1 + \frac{4}{9}} = \frac{{\sqrt {13} }}{3}$.

Here $\Delta = 0$.

==> $2b = 5$ and $2ae = 13$.

Now, also we know for hyperbola

${b^2} = {a^2}({e^2} - 1)$

==> $\frac{{25}}{4} = \frac{{{{(13)}^2}}}{{4{e^2}}}({e^2} - 1)$

==> $\frac{{25}}{4} = \frac{{169}}{4} - \frac{{169}}{{4{e^2}}}$ or ${e^2} = \frac{{169}}{{144}}$

==> $e = \frac{{13}}{{12}}$

or $a = 6,\,b = \frac{5}{2}$ or hyperbola is $\frac{{{x^2}}}{{36}} - \frac{{{y^2}}}{{25/4}} = 1$

==> $25{x^2} - 144{y^2} = 900$.

Also $(5, -2)$ satisfies it, so $\frac{4}{{49}}(25) - \frac{{51}}{{196}}(4) = 1$

and ${a^2} = \frac{{49}}{4}$

==> $a = \frac{7}{2}$.

==> $\frac{{16{{(x + 2)}^2}}}{{16}} - \frac{{{{(y - 2)}^2}}}{{16}} = 1$

Transverse and conjugate axes are $y = 2$, $x = - 2$.

==> $a = 4$and focus $(6,0)$

==> $ae = 4$

==> $e = \frac{3}{2}$.

Therefore $b = 2\sqrt5$

Hence required equation is $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{{20}} = 1$

$i.e.$, $5{x^2} - 4{y^2} = 80$.

But it passes through $(3, 2)$

==> $\frac{9}{{{a^2}}} - \frac{4}{{{b^2}}} = 1$.....$(i)$

Also its passes through $(-17, 12)$

==> $\frac{{{{( - 17)}^2}}}{{{a^2}}} - \frac{{{{(12)}^2}}}{{{b^2}}} = 1$ .....$(ii)$

Solving these, we get $a = 1$ and $b = \sqrt 2 $

Hence length of transverse axis $ = 2a = 2$.

$\frac{x}{a} + \frac{y}{b} = \frac{1}{m}$ .....$(ii)$

Multiplying equation $(i)$ and $(ii),$

$\left( {\frac{x}{a} - \frac{y}{b}} \right)\,\left( {\frac{x}{a} + \frac{y}{b}} \right) = m.\frac{1}{m}$

$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$, which is the equation of hyperbola.

Hence, $a = \frac{3}{{\sqrt 3 }} = \sqrt 3 $

Therefore, equation is $\frac{{{x^2}}}{3} - \frac{{{y^2}}}{9} = 1$

$i.e.$, $3{x^2} - {y^2} = 9$.

==> $5[{x^2} + {y^2} - 4x - 2y + 5]$

$ = 4[{x^2} + 4{y^2} + 1 + 4xy - 2x - 4y]$

==> ${x^2} - 11{y^2} - 16xy - 12x + 6y + 21 = 0$.

$ae - a = 8$ or $e = 1 + \frac{8}{5} = \frac{{13}}{5}$

$b = 5\sqrt {\frac{{{{13}^2}}}{{{5^2}}} - 1} = 5 \times \frac{{12}}{5} = 12$

and centre of hyperbola$ \equiv (5,\,0)$   

$\frac{{{{(x - 5)}^2}}}{{{5^2}}} - \frac{{{{(y - 0)}^2}}}{{{{12}^2}}} = 1$.

$⇒$  $\frac{9}{{16}} = ({e^2} - 1)$

$⇒$  $e = \frac{5}{4}$

Vertex is $(0, 2)$.

Hence focus is $( \pm ae,\,2) = ( \pm 5,\,2)$.

$\left( {\frac{{hf - bg}}{{ab - {h^2}}},\,\frac{{gh - af}}{{ab - {h^2}}}} \right)$

$= \left( {\frac{{ + 16.9}}{{ - 9.16}},\,\frac{{ - 9(16)}}{{ - 9(16)}}} \right) = ( - 1,\,1)$.

$e = 2$ and centre is $\left( {\frac{{6 - 4}}{2},4} \right) = (1,4)$

$⇒$ $6 = 1 + ae$

$⇒$ $ae = 5$

$⇒$ $a = \frac{5}{2}$and $b = \frac{5}{2}(\sqrt 3 )$

Hence the required equation is $\frac{{{{(x - 1)}^2}}}{{(25/4)}} - \frac{{{{(y - 4)}^2}}}{{(75/4)}} = 1$

or $12{x^2} - 4{y^2} - 24x + 32y - 127 = 0$.

$ \Rightarrow \frac{{{{(x - 1)}^2}}}{{16}} - \frac{{{{(y + 1)}^2}}}{9} = 1$

==> Latus rectum $ = \frac{{2{b^2}}}{a} = \frac{{2 \times 9}}{4} = \frac{9}{2}$.

$5{(x + 2)^2} - 4\,{(y - 1)^2} = 20$ ==> $\frac{{{{(x + 2)}^2}}}{4} - \frac{{{{(y - 1)}^2}}}{5} = 1$

From ${b^2} = {a^2}({e^2} - 1)$,   $5 = 4({e^2} - 1)$

$ \Rightarrow {e^2} = 9/4 $

$\Rightarrow e = 3/2$.

==> $\frac{{dx}}{{d\phi }} = 2\sec \phi \tan \phi $

Differentiate, $y = 3\tan \phi $  w.r.t. $\theta$,

we get $\frac{{dy}}{{d\phi }} = 3{\sec ^2}\phi $

$\therefore $Gradient of tangent $\frac{{dy}}{{dx}} = \frac{{dy/d\phi }}{{dx/d\phi }} = \frac{{3{{\sec }^2}\phi }}{{2\sec \phi \tan \phi }}$

$\frac{{dy}}{{dx}} = \frac{3}{2}\,{\rm{cosec}}\phi $ .....$(i)$

But, tangent is parallel to $3x - y + 4 = 0$

$\therefore $Gradient $ m = 3$ .....$(ii)$

By $(i)$ and $(ii),$ $\frac{3}{2}{\rm{cosec}}\phi = 3$

==> ${\rm{cosec}}\phi = 2$,

$\therefore \phi = 30^\circ $.

$\frac{{{a^2}x}}{{a\sec \theta }} + \frac{{{b^2}y}}{{b\tan \theta }} = {a^2} + {b^2}$.

$\frac{{ax}}{{\sec \theta }} + \frac{{by}}{{\tan \theta }} = {a^2} + {b^2}$ .....$(i)$

But it is given by $lx + my - n = 0$.....$(ii)$

Comparing $(i)$ and $(ii),$ we get

$\sec \theta = \frac{a}{l}\left( {\frac{{ - n}}{{{a^2} + {b^2}}}} \right)$

and $\tan \theta = \frac{b}{m}\left( {\frac{{ - n}}{{{a^2} + {b^2}}}} \right)$

Hence eliminating $\theta $, we get

$\frac{{{a^2}}}{{{l^2}}} - \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} + {b^2})}^2}}}{{{n^2}}}$.

$\frac{{16x}}{8} + \frac{{9y}}{{3\sqrt 3 }} = 16 + 9\,\,$

$i.e.,$$2x + \sqrt 3 y = 25$

Trick : This is the only equation among the given options at which the point $(8,\,3\sqrt 3 )$ is located.

$\frac{{{a^2}(x - {x_1})}}{{{x_1}}} = \frac{{{b^2}(y - {y_1})}}{{ - {y_1}}}$

Here, ${a^2} = 267,\,{b^2} = 48$ and $({x_1},{y_1}) = (6,4)$

$\therefore \frac{{27(x - 6)}}{6} = - \frac{{48(y - 4)}}{4}$

==> $3(x - 6) = - 8(y - 4)$

==> $3x + 8y = 50$.

$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ at point $(a\sec \theta ,\,b\tan \theta )$

is $ax\sec \theta + by\cot \theta = {a^2} + {b^2}$

or $y = \frac{{ - a}}{b}\sin \theta \,x + \frac{{{a^2} + {b^2}}}{{b\cot \theta }}$

Comparing above equation with equation $y = mx + \frac{{25\sqrt 3 }}{3}$ and taking $a = 4,\,b = 3$

we get, $\frac{{{a^2} + {b^2}}}{{b\cot \theta }} = \frac{{25\sqrt 3 }}{3}$

$ \Rightarrow $ $\tan \theta = \sqrt 3$

$\Rightarrow \theta = {60^o}$

and $m = - \frac{a}{b}\sin \theta = \frac{{ - 4}}{3}\sin {60^o}$

= $\frac{{ - 4}}{3} \times \frac{{\sqrt 3 }}{2} = \frac{{ - 2}}{{\sqrt 3 }}$.

$ \Rightarrow $$\frac{{2x}}{{16}} - \frac{{2y}}{9}\frac{{dy}}{{dx}} = 0$

==> $\frac{{dy}}{{dx}} = \frac{{2x \times 9}}{{16 \times 2y}}$

$ = \frac{9}{{16}}\frac{x}{y}$

==> ${\left( {\frac{{ - dx}}{{dy}}} \right)_{( - 4,0)}}$

$= \frac{{ - 16}}{9}\frac{y}{x} = 0$

Hence, equation of normal

==> $(y - 0)\, = 0(x + 4)$

==> $y = 0.$

Then its conjugate will be, $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = - 1$.....$(ii)$

If $e$ is eccentricity of hyperbola $(i),$ then ${b^2} = {a^2}({e^2} - 1)$

or $\frac{1}{{{e^2}}} = \frac{{{a^2}}}{{({a^2} + {b^2})}}$ .....$(iii)$

Similarly if e' is eccentricity of conjugate $(ii),$ then ${a^2} = {b^2}(e{'^2} - 1)$ or

$\frac{1}{{e{'^2}}} = \frac{{{b^2}}}{{({a^2} + {b^2})}}$.....$(iv)$

Adding $(iii)$ and $(iv),$

$\frac{1}{{{{(e')}^2}}} + \frac{1}{{{e^2}}}$

$= \frac{{{a^2}}}{{{a^2} + {b^2}}} + \frac{{{b^2}}}{{{a^2} + {b^2}}} = 1.$

Product of length of perpendiculars drawn from any point on the hyperbola $(i)$ to the asymptotes is

$\frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}}= \frac{{2 \times 1}}{{2 + 1}} = \frac{2}{3}$.

==> $e = \frac{4}{5}$

Therefore, focus is $(-4, 0), (4, 0).$

Given eccentricity of hyperbola = $2$

$a = \frac{{ae}}{e} = \frac{4}{2} = 2$ and $b = 2\sqrt {(4 - 1)} = 2\sqrt 3 $

Hence hyperbola is $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{{12}} = 1$.

Here, co-ordinates of focus are

$(ae\cos 45^\circ ,\,ae\sin 45^\circ ) \equiv (c\sqrt 2 ,c\sqrt 2 )$, 

$\{ \because e = \sqrt 2 $,$a = c\sqrt 2 \} $

Similarly other focus is $( - c\sqrt 2 , - c\sqrt 2 )$

$\frac{x}{{(a/\sec \theta )}} - \frac{y}{{(b/\tan \theta )}} = 1$

or $\frac{a}{{\sec \theta }} = 1,\,\,\frac{b}{{\tan \theta }} = 1$

==> $a = \sec \theta $, $b = \tan \theta $

or $(a,b)$ lies on ${x^2} - {y^2} = 1$.

$\therefore $ $ee'\, = 1.$

==> ${x^2} - 2x - 3{y^2} = 8$

==> ${(x - 1)^2} - 3{y^2} = 9$

==> $\frac{{{{(x - 1)}^2}}}{9} - \frac{{{y^2}}}{3} = 1$

Conjugate of this hyperbola is $ - \frac{{{{(x - 1)}^2}}}{9} + \frac{{{y^2}}}{3} = 1$

and its eccentricity $(e) = \sqrt {\left( {\frac{{{a^2} + {b^2}}}{{{b^2}}}} \right)} $

Here, ${a^2} = 9$, ${b^2} = 3$;

$\therefore $ $e = \sqrt {\frac{{9 + 3}}{3}} = 2$.

==> $2{b^2} = 9a$…..$(i)$

Now ${b^2} = {a^2}({e^2} - 1) = \frac{9}{{16}}{a^2}$

==> $a = \frac{4}{3}b$…..$(ii),$  ($e = \frac{5}{4}$)

From $(i)$ and $(ii),$ $b = 6$, $a = 8$

Hence, equation of hyperbola $\frac{{{x^2}}}{{64}} - \frac{{{y^2}}}{{36}} = 1$.

Equation of tangent are equally inclined to the axis

$i.e.$, $\tan \theta = 1 = m$.

 Eq. of tangent $y = mx + \sqrt {{a^2}{m^2} - {b^2}} $

Given eq. $\frac{{{x^2}}}{3} - \frac{{{y^2}}}{2} = 1$ is a eq. of hyperbola

which is of form $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$.

Now, on comparing ${a^2} = 3$, ${b^2} = 2$

 $y = 1.x + \sqrt {3 \times {{(1)}^2} - 2} $

==> $y = x + 1$.
 

hyperbola $3{x^2} - 2{y^2} + 4x - 6y = 0$

$\therefore$ Equation of the chord is $T = {S_1}$

i.e. $3x{x_1} - 2y{y_1} + 2(x + {x_1}) - 3(y + {y_1}) = 0$

or $(3{x_1} + 2)x - (2{y_1} + 3)y + (2{x_1} - 3{y_1}) = 0$

If this chord is parallel to line $y = 2x,$ then

${m_1} = {m_2}$==> $ - \frac{{3{x_1} + 2}}{{ - (2{y_1} + 3)}} = 2$ ==>$3{x_1} - 4{y_1} = 4$

Hence the locus of the middle point $({x_1},{y_1})$ is $3x -4y=4$.

$({x_1},\,{y_1})$ be any point on hyperbola, $\therefore \,\frac{{x_1^2}}{{{a^2}}} - \frac{{y_1^2}}{{{b^2}}} = 1$ or

${b^2}x_1^2 - {a^2}y_1^2 = {a^2}{b^2}$.

We also know that asymptotes of given hyperbola are

$\frac{{{x^2}}}{{{a^2}}}$$ - \frac{{{y^2}}}{{{b^2}}} = 0$

$\therefore \,$ Product of $ \bot $ from $({x_1},{y_1})$ to pair of lines $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 0$ is

$\frac{{|Ax_1^2 + 2H{x_1}{y_1} + By_1^2|}}{{\sqrt {{{(A - B)}^2} + 4{H^2}} }}$; $\frac{{{b^2}x_1^2 - {a^2}y_1^2}}{{\sqrt {{{({b^2} + {a^2})}^2}} }} = \frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}}$.

Point of intersection of $y=x^2+2 a x+2021$ and $y=0$ is

$x ^2+2 ax +2021=0$, has rational roots

$D =4 a ^2-4 \times 2021=4\left( a ^2-2021\right)$

$\Rightarrow a ^2-2021$ is perfect square

$a^2-2021=\lambda^2 \Rightarrow(a-\lambda)(a+\lambda)=2021$

$(a-\lambda)(a+\lambda)=1.2021$

$=43.47$

$=47.43$

$=2021.1$

for max,

$a+\lambda=2021$

$a-\lambda=1$

$a=1011, \lambda=1010$

max value of $a=1011$

$F _1 F _2= OF _1=2 ae$

$OC =\sqrt{ a ^2+ b ^2}$

$\cos 120^{\circ}=\frac{( ae )^2+4( ae )^2- a ^2- b ^2}{2 \cdot ae \cdot 2 ae }$

$\Rightarrow-\frac{1}{2}=\frac{5( ae )^2- a ^2- a ^2\left(1- e ^2\right)}{4 a ^2 e ^2}$

$\Rightarrow 8 e ^2=2$

$e =\frac{1}{2}$

Here, $\angle F_1 C B=90^{\circ}$

$\frac{-b}{a} \cdot \frac{b}{a e}=-1 \Rightarrow \frac{b^2}{a^2}=e \Rightarrow e=1-e^2$

$e^2+e-1=0$

$e=\frac{-1 \pm \sqrt{5}}{2} \stackrel{\overline{5}}{2} \Rightarrow e=\frac{\sqrt{5}-1}{}$

Equation of parabola

$y=x^2-4$ and $2 y=4-x^2$

Let point $B=\left(h, \frac{4-h^2}{2}\right)$

$A=\left(-h, \frac{4-h^2}{2}\right)$

$C=\left(h_1 h^2-4\right)$

$D=\left(-h, h^2-4\right)$

Area of rectangle

$A B C D=A B \times B C=2 h \times\left(\frac{4-h^2}{2}-h^2+4\right)$

$\Rightarrow A=12 h-3 h^3 \Rightarrow \frac{d A}{d h}=12-9 h^2$

For maxima or minima

$\frac{d A}{d h}=0 \Rightarrow 12-9 h^2=0$

$h=\pm \frac{2}{\sqrt{3}}$ maximum at $h=\frac{2}{\sqrt{3}}$

$A=12\left(\frac{2}{\sqrt{3}}\right)-3\left(\frac{2}{\sqrt{3}}\right)^3=\frac{24}{\sqrt{3}}-\frac{8}{\sqrt{3}}$

$=\frac{16}{\sqrt{3}}=\frac{16 \sqrt{3}}{3}$

$A-\frac{16 \times 173}{3}-9.22$

${[A]=[9.22]=9 }$

We have,

Equation of ellipse

$\frac{x^2}{9}+\frac{y^2}{4}=1$

Equation of $\operatorname{tangent}$ at $(3 \cos \theta, 2 \sin \theta)$ is

$\frac{x}{3} \cos \theta+\frac{y}{2} \sin \theta=1$

Intercept of tangent is $(3 \sec \theta, 0)$ and $(0,2 \operatorname{cosec} \theta)$.

Area of quadrilateral

$=4 \times \cdot \frac{1}{2} 3 \sec \theta \times 2 \operatorname{cosec} \theta$

$=12 \sec \theta \operatorname{cosec} \theta=\frac{24}{\sin 2 \theta}$

Minimum area $=24$

Given, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$A B C$ is an equilateral traingles.

$\ln \Delta A O F_2^1$

$\tan 30^{\circ}=\frac{O F_2}{O A}$

$\frac{1}{\sqrt{3}} =\frac{a e}{b}$

$b =\sqrt{3} a e$

$b^2 =3 a^2 e^2$

$\frac{b^2}{a^2} =3 e^2$

$4 e^2 =1$

$e^2 =\frac{1}{4} \Rightarrow e=\frac{1}{2}$

Given, $x^2=4 k y$

$B C$ is latusrectum.

$B C=4 k$

$B D=D E=E C$

$D E=\frac{B C}{3}=\frac{4 k}{3}$

$P$ is centre of ellipse.

$P E=\frac{2 k}{3}$

$O P^2=k$

$\because$ Eccentricity of ellipse

$\sqrt{1-\frac{P E^2}{O P^2}}=\sqrt{1-\frac{4 k^2}{9 k^2}}$

$e=\sqrt{\frac{9-4}{9}}=\frac{\sqrt{5}}{3}$

We have rectangular hyperbola

$x^2-y^2=a^2$

Given $A B C$ is an equilateral triangle.

$A B =B C=A C$

$A B^2 =B C^2$

$a^2(\sec \theta+1)^2+a^2 \tan ^2 \theta=4 a^2 \tan ^2 \theta$

$\begin{aligned}(\sec \theta+1)^2 &=3 \tan ^2 \theta \$\sec \theta+1)^2 &=3\left(\sec ^2 \theta-1\right) \$\sec \theta+1)^2 &=3(\sec \theta+1)(\sec \theta-1) \end{aligned}$

$\sec \theta+1=3 \sec \theta-3$

$\sec \theta=2$

$\theta=60^{\circ}$

$\because$ Side $B C=2 a \tan \theta$

$=2 a \tan 6 \theta^{\circ}=2 a \sqrt{3}$

But side of triangle is $k a$.

$k a =2 a \sqrt{3}$

$k =2 \sqrt{3}$

Hence, $k \in(2,4]$.

An ellipse passes through $(0,0)$, $(1,0)$ and $(2,0)$. Its minor and major axis are parallel to coordinates axes. One of its foci lies on $Y$-axis.

Let one foci is $(0, k)$.

$\therefore O B$ is the latusrectum of ellipse $F_1(0, k)$ is mid-point of $O B$

$F_1(0,1) \quad\left[\because k=\frac{2+0}{2}=1\right]$

Now $F_2(h, k)=F_2(h, 1)$

Now by definition of ellipse

$B F_1+B F_2=2 a =O F_1+O F_2$

$\therefore \quad B F_1+B F_2 =O F_1+O F_24$

$\sqrt{1+1}+\sqrt{(h-1)^2+1} =\sqrt{0+1}+\sqrt{h^2+1}$

$\sqrt{2}+\sqrt{(h-1)^2+1} =1+\sqrt{h^2+1}$

$\Rightarrow \quad h =1$

$\therefore F_1(0,1) \text { and } F_2(1,1)$

$F_1 F_2=2 a e =1$

$\therefore \quad O F_1+O F_2=2 a =1+\sqrt{2}$

$\Rightarrow \quad e=\frac{1}{\sqrt{2}+1} \Rightarrow e =\frac{\sqrt{2}-1}{2-1}$

$e =\sqrt{2}-1$

Equation of ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a > b$

Equation of parabola $x^2=4(y+b)$

Foci of ellipse $(\pm a e, 0)$.

End of latusrectum of parabola

$=(\pm 2,1-b)$

$A B C D$ is a square

$A B=C D \Rightarrow 2 a e=4 \Rightarrow a e=2$

$B C=C D \Rightarrow B C^2=C D^2$

$(2-a e)^2+(1-b)^2=4^2$

$(2-a e)^2+(1-b)^2=4^2$

$0+(1-b)^2=4^2$

$1-b=\pm 4$

$b=5, b=-3 \Rightarrow a=\sqrt{29}$ or $\sqrt{13}$

$e=\sqrt{1-\frac{b^2}{a^2}}$

$e=\sqrt{1-\frac{9}{13}}=\frac{2}{\sqrt{13}}$ or $2$

Let the position of cannon $=P$ speed of sound $=S$

Time of heard of sound of cannon firing of at $A=t$

$\therefore$ at $B=t+1$

Hence, $P A=$ Speed $\times$ Time

$P A=S t$

$and\,P B=S(t+1)=S t+S$

$P B-P A=S t+S-S t$

$P B-P A=S$

$P B-P A=$ constant

$\therefore A$ and $B$ are foci of hyperbola.

Given parabola $(y-k)^2=4(x-h)$

This is passes through $(0,0)$ and $(0,2)$.

$\therefore k^2=-4\,h$

and $(2-k)^2=-4\,h$

From Eqs. $(i)$ and $(ii)$, we get

$k=1, h=-\frac{1}{4}$

$\therefore$ Equation of parabola becomes

$(y-1)^2=4 x+1$

Focus of parabola $=\left(\frac{3}{4}, 1\right)$

End of latusrectum $D=\left(\frac{3}{4}, 3\right)$

Axis of parabola is $y-1=0$

Point $P$ intersection of axis of parabola and $Y$-axis.

$P(0,1)$

$D\left(\frac{3}{4}, 3\right)$

$A\left(-\frac{1}{4}, 1\right) \quad[\because A$ is vertex of parabola $]$ Slope of $P D=\frac{3-1}{3 / 4-0}=8 / 3=m_1$

Slope of $D A=\frac{3-1}{\frac{3}{4}+\frac{1}{4}}=2=m_2$

$\therefore \quad \angle P D A=\theta=\tan ^{-1}\left(\frac{m_1-m_2}{1+m_1 m_2}\right)$

$=\tan ^{-1}\left(\frac{8 / 3-2}{1+\frac{16}{3}}\right)=\tan ^{-1}\left(\frac{2}{19}\right)$

Mid-point $(h, k)=\left(\frac{x-x_1}{2}, \frac{x_1}{2}\right)$

$A B =l$

$A B^2 =l^2$

$\Rightarrow\left(x+x_1\right)^2+x_1^2=l^2$

$\frac{x-x_1}{2} =h \Rightarrow x-x_1=2 h$

$\frac{x_1}{2} =k \Rightarrow x_1=2 k$

$x+x_1 =2 h+4 k$

Hence, locus of mid-point $(h, k)$ is

$(2 h+4 k)^2+4 k^2 =l^2$

$\Rightarrow \quad x^2+5 y^2+4 x y =\frac{l^2}{4}$

$\text { Area of ellipse } =\frac{\pi l^2}{4}$

Given,

$B O A C$ is a rectangle in the $X Y$-plane where $O$ is the origin and $A, B$ lie on $y=x^2$

Point $A=\left(t_1, t_1^2\right)$

$B=\left(t_2, t_2^2\right)$

Slope of $O A=t_1$

Slope of $O B=t_2$

$O A$ is perpendicular to $O B$.

$\therefore \quad t_1 t_2=-1$

$h=t_1+t_2$ and $k=t_1^2+t_2^2$

$k=\left(t_1+t_2\right)^2-2 t_1 t_2$

$k=h^2+2$

$\therefore$ Locus of $C$ is $y=x^2+2$

We have, $y=a x^2+b x+c$

Parabola has two roots one is positive and one is negative.

Clearly, $c$ is negative

$a > 0$

$-\frac{b}{a} =2$

$\therefore \quad-\frac{b}{a} > 0$

$\therefore \quad b < 0$

$\therefore > 0$

Hence, option $(b)$ is correct.

Given, in cllipse

$A^{\prime} S^{\prime}=S S^{\prime}=S A$

$\therefore \quad S S^{\prime} =\frac{1}{3} A A^{\prime}$

$\Rightarrow \quad 2 a e =\frac{1}{3}(2 a)$

$\Rightarrow \quad =\frac{1}{3}$

$\text { Also given, } b =2 \sqrt{2}$

$\Rightarrow \quad \frac{1}{9} =1-\frac{b^2}{a^2}$

$\Rightarrow \quad \frac{8}{a^2} =1-\frac{1}{9}=\frac{8}{9}$

$\Rightarrow \quad a^2 =9$

$\therefore \quad a =3$

Let equation of ellipse is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\therefore$ Equation of circle is

$x^2+(y+b)^2=r^2$

Put $x^2=a^2-\frac{a^2 y^2}{b^2}$ in circle

$a^2-\frac{a^2 y^2}{b^2}+(y+b)^2=r^2$

$\Rightarrow\left(1-\frac{a^2}{b^2}\right) y^2+2 b y+\left(a^2+b^2-r^2\right)=0$

$D=0 \Rightarrow r^2=\frac{a^4}{a^2-b^2} \Rightarrow b=a \sqrt{1-\frac{a^2}{r^2}}$

Area of ellipse $=\pi a b$

$A=\pi a^2 \sqrt{1-\frac{a^2}{r^2}}$

$\frac{d A}{d a}=0 \Rightarrow a^2=\frac{2 r^2}{3} \Rightarrow a=\sqrt{\frac{2}{3} r}$

$\therefore \quad b=a \sqrt{1-\frac{2}{3}}=\frac{a}{\sqrt{3}}$

$\Rightarrow \quad e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{1}{3}}=\sqrt{\frac{2}{3}}$

We have,

Equation of parabola

On solving Eqs. $(i)$ and $(ii)$, we get $(1,-2)$ and $(4,4)$

Now, $A B C$ is an isosceles triangle $A B=A C$

$\therefore \quad(x-1)^2+(2)^2=(x-4)^2+(4)^2$

$\Rightarrow x^2-2 x+1+4=x^2-8 x+16+16$

$\Rightarrow \quad 6 x=27$

$\Rightarrow \quad x=\frac{27}{6}=\frac{9}{2}$

$\therefore \quad$ Coordinate $=\left(\frac{9}{2}, 0\right) .$

We have,

$e x^2+\pi y^2-2 e^2 x-2 \pi^2 y+e^3+\pi^3=\pi e$

$\Rightarrow e x^2-2 e^2 x+\pi y^2-2 \pi^2 y=\pi e-e^3-\pi^3$

$\Rightarrow e\left(x^2-2 e x+e^2\right)+\pi\left(y^2-2 \pi y+\pi^2\right)$

$=\pi e-e^3-\pi^3+e^3+\pi^3$

$\Rightarrow e(x-e)^2+\pi(y-\pi)^2=\pi e$

$\Rightarrow \frac{(x-e)^2}{(\sqrt{\pi})^2}+\frac{(y-\pi)^2}{(\sqrt{e})^2}=1$

$\therefore \quad \sqrt{\pi}>\sqrt{e}$

$\therefore \quad P S_1+P S_2=2 a$

$\therefore \quad P S_1+P S_2=2 \sqrt{\pi}$

We have,

$4 x^2+9 y^2-8 x-36 y+15=0$

$\Rightarrow 4\left(x^2-2 x+1\right)+9\left(y^2-4 y+4\right)$

$=-15+4+36$

$\Rightarrow \quad 4(x-1)^2+9(y-2)^2=25$

$\Rightarrow \quad \frac{(x-1)^2}{25 / 4}+\frac{(y-2)^2}{25 / 9}=1$

Now, $\min \left(x^2-2 x+y^2-4 y+5\right)$

$+\max \left(x^2-2 x+y^2-4 y+5\right)$

$=\min \left[(x-1)^2+(y-2)^2\right]$

$+\max \left[(x-1)^2+(y-2)^2\right]$

$=\frac{25}{9}+\frac{25}{4}$

$=25\left(\frac{4+9}{36}\right)=\frac{25 \times 13}{36}=\frac{325}{36}$

We have, $y^2=6 x$

Equation of normal of parabola $y^2=4 a x$ is $y=m x-2 a m-a m^3$

$\therefore$ Equation of normal of parabola $y^2=6 x$ is

$y=m x-3 m-\frac{3}{2} m^3$

Normal is passes through $(\lambda, 0)$.

$\therefore \quad 0=\lambda m-3 m-\frac{3}{2} m^3$

$\Rightarrow \quad \lambda=3+\frac{3}{2} m^2 \Rightarrow m^2=\frac{2}{3}(\lambda-3)$

$\Rightarrow \quad m=\sqrt{\frac{2}{3}(\lambda-3)}$

$\therefore m$ is real.

$\therefore \quad \lambda-3 > 0 \Rightarrow \lambda > 3$

We have equation of ellipse

$9 x^2+16 y^2 =144$

$\Rightarrow \quad \frac{x^2}{16}+\frac{y^2}{9} =1$

Equation of tangent of ellipse is

$y =m x \pm \sqrt{a^2 m^2+b^2}$

$\therefore \quad y =m x \pm \sqrt{16 m^2+9}$

Now, given $y$-intercept $=5$

$\therefore \sqrt{16 m^2+9}=5 \Rightarrow 16 m^2+9=25$

$\Rightarrow \quad 16 m^2=16 \Rightarrow m=\pm 1$

$\therefore \text { Positive slope }=1$

We have,

$(y-a x-b)\left(b x^2+a y^2-a b\right)=0$

$\Rightarrow y-a x-b=0 \text { or } b x^2+a y^2-a b=0$

$\Rightarrow y=a x+b \text { or } b x^2+a y^2=a b$

$\Rightarrow y=a x+b \text { or } \frac{x^2}{a}+\frac{y^2}{b}=1$

Case $I$ When $a,b > 0$

We have,

Equation of parabola

$y^2=4 x$

Equation of tangent of parabola at $(1,2)$, is

$2 y=4 \frac{(x+1)}{2} \Rightarrow y=x+1$

The tangent of parabola $y=x+1$ intersect the dircctrix of parabola at $A(-1,0)$

Now, $A B$ and $A C$ are tangent of circle

$\begin{array}{lc}\because & A B=A C \\ \Rightarrow \sqrt{(1+1)^2+(2-0)^2}\end{array}$

$=\sqrt{(-1+1)^2+(0-k)^2}$

$\Rightarrow \quad 4+4=k^2 \Rightarrow k^2=8 \Rightarrow k=2 \sqrt{2}$

We have, equation of ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Point $P(a \cos \theta, b \sin \theta)$ lie on ellipse

Foci of ellipse $\left(F_1\right)(a e, 0)$ and $F_2(-a e, 0)$