Maths Solve the Following Question.(4 Marks)

Let $\theta_1$ and $\theta_2$ be the inclinations.

$m _1=\tan \theta_1, m _2=\tan \theta_2$

Let $P\left(x_1, y_1\right)$ be a point on the hyperbola

Equation of a tangent with slope ' $m$ ' to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is

$y = mx \pm \sqrt{a^2 m^2-b^2}$

This tangent passes through $P \left( x _1, y _1\right)$.

$\begin{aligned} & \left( y _1- mx_{1 } \right)^2=a^2 m ^2- b ^2 \\ & \left(x_1^2- a ^2\right) m ^2-2 x_1 y_1 m +\left(y_1^2+ b ^2\right)=0 \ldots \ldots \text { (i) } \\ & \end{aligned}$

This is a quadratic equation in ‘m’. It has two roots say m1 and m2, which are the slopes of two tangents drawn from P.

$\therefore m _1+ m _2=\frac{2 x_1 y_1}{x_1^2-a^2}$

$\begin{aligned} & \text { Since } \tan \theta_1+\tan \theta_2=k, \\ & \frac{2 x_1 y_1}{x_1^2-a^2}=k\end{aligned}$

$\therefore P \left( x _1, y _1\right)$ moves on the curve whose equation is $k \left( x ^2- a ^2\right)=2 xy$.

$\therefore \frac{x^2}{6}-\frac{y^2}{8}=1$

Comparing this equation with $\frac{x^2}{ a ^2}-\frac{y^2}{h^2}=1$, we get

$a^2=6$ and $b^2=8$

Given equation of line is 2x – y = 4

∴ y = 2x – 4

Comparing this equation with y = mx + c, we get

m = 2 and c = -4

For the line $y=m x+c$ to be a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, we must have

$\begin{aligned} & c^2=a^2 m^2-b^2 \\ & c^2=(-4)^2=16 \\ & a^2 m^2-b^2=6(2)^2-8=24-8=16\end{aligned}$

∴ The given line is a tangent to the given hyperbola and point of contact

$\begin{aligned} & =\left(-\frac{ a ^2 m }{ c },-\frac{ b ^2}{ c }\right) \\ & =\left(\frac{-6(2)}{-4}, \frac{-8}{-4}\right) \\ & =(3,2)\end{aligned}$

$\therefore \frac{x^2}{5}+\frac{y^2}{4}=1$

Comparing this equation with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we get

$a^2=5$ and $b^2=4$

Since inclinations of tangents are $\theta_1$ and $\theta_2$,

$m_1=\tan \theta_1$ and $m_2=\tan \theta_2$

Equation of tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{h^2}=1$ having slope $m$ are

$\begin{aligned} & y=m x \pm \sqrt{a^2 m^2+b^2} \\ & \therefore y=m x \pm \sqrt{5 m^2+4} \\ & \therefore y-m x= \pm \sqrt{5 m^2+4}\end{aligned}$

Squaring both the sides, we get

$\begin{aligned} & y=m x \pm \sqrt{a^2 m^2+b^2} \\ & \therefore y=m x \pm \sqrt{5 m^2+4} \\ & \therefore y-m x= \pm \sqrt{5 m^2+4}\end{aligned}$

Squaring both the sides, we get

$\begin{aligned} & y^2-2 m x y+m^2 x^2=5 m^2+4 \\ & \therefore\left(x^2-5\right) m^2-2 x y m+\left(y^2-4\right)=0\end{aligned}$

The roots m1 and m2 of this quadratic equation are the slopes of the tangents.

$\therefore m _1+ m _2=\frac{-(-2 x y)}{x^2-5}=\frac{2 x y}{x^2-5}$

Given, $\tan \theta_1+\tan \theta_2=2$

$\begin{aligned} & \therefore m _1+ m _2=2 \\ & \therefore \frac{2 x y}{x^2-5} \\ & \therefore xy = x ^2-5 \\ & \therefore x ^2- xy -5=0, \text { which is the required equation of the locus of } P .\end{aligned}$

$\therefore \frac{x^2}{17}+\frac{y^2}{\frac{17}{4}}=1$

Comparing this equation with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ we get

$a^2=17$ and $b^2=\frac{17}{4}$

Given equation of line is $8 y+x=17$,

i.e., $y=\frac{-1}{8} x+\frac{17}{8}$

Comparing this equation with $y=m x+c$, we get

$m=\frac{-1}{8}$ and $c=\frac{17}{8}$

For the line $y = mx + c$ to be a tangent to the ellipse $\frac{x^2}{ a ^2}+\frac{y^2}{ b ^2}=1$, we must have

$\begin{aligned} & c^2=a^2 m^2+b^2 \\ & c^2=\left(\frac{17}{8}\right)^7=\frac{289}{64} \\ & a^2 m^2+b^2=17\left(\frac{-1}{8}\right)^2+\frac{17}{4} \\ & =\frac{17}{64}+\frac{17}{4} \\ & =\frac{289}{64} \\ & =c^2\end{aligned}$

$\therefore$ The given line touches the given ellipse and point of contact is

$\begin{aligned} & \left(\frac{-a^2 m}{c}, \frac{b^2}{c}\right)=\left(\frac{-17\left(\frac{-1}{8}\right)}{\frac{17}{8}}, \frac{\frac{17}{4}}{\frac{17}{8}}\right) \\ & =(1,2) .\end{aligned}$

Comparing this equation with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we get

$a^2=5$ and $b^2=4$

Equations of tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ having slope $m$ are

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

Since (2, -2) lies on both the tangents,

$\begin{aligned} & -2=2 m \pm \sqrt{5 m^2+4} \\ & \therefore-2-2 m= \pm \sqrt{5 m^2+4}\end{aligned}$

Squaring both the sides, we get

$\begin{aligned} & 4 m ^2+8 m+4=5 m ^2+4 \\ & \therefore m ^2-8 m =0 \\ & \therefore m ( m -8)=0 \\ & \therefore m =0 \text { or } m =8\end{aligned}$

These are the slopes of the required tangents.

∴ By slope point form y – y1 = m(x – x1),

the equations of the tangents are y + 2 = 0(x – 2) and y + 2 = 8(x – 2)

∴ y + 2 = 0 and y + 2 = 8x – 16

∴ y + 2 = 0 and 8x – y – 18 = 0.

Comparing this equation with $\frac{x^2}{ a ^2}-\frac{y^2}{ b ^2}=1$, we get

$\begin{aligned} & a ^2=100 \text { and } b ^2=25 \\ & \therefore a =10 \text { and } b=5 \\ & \therefore \text { Co-ordinates of vertex is } A(a, 0) \text {, i.e., } A(10,0)\end{aligned}$

$\begin{aligned} & \text { Eccentricity, } e=\frac{\sqrt{a^2+b^2}}{a} \\ & =\frac{\sqrt{100+25}}{10} \\ & =\frac{\sqrt{125}}{10} \\ & =\frac{5 \sqrt{5}}{10} \\ & =\frac{\sqrt{5}}{2}\end{aligned}$

Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0)

i.e., $S\left(10\left(\frac{\sqrt{5}}{2}\right), 0\right)$ and $S^{\prime}\left(-10\left(\frac{\sqrt{5}}{2}\right), 0\right)$

i.e., S(5√5, 0) and S'(-5√5, 0)

Since S, A and S’ lie on the X-axis,

SA = |5√5 – 10| and S’A = |-5√5 – 10|

= |-(5√5 + 10)|

= |5√5 + 10|

∴ SA . S’A = |5√5 – 10| |5√5 + 10|

= |(5√5)2 – (10)2|

= |125 – 100|

= |25|

SA . S’A = 25

Let $P\left(x_1, y_1\right)$ be the point of contact.
$\therefore x_1=\frac{-a^2 m}{c}=\frac{-25\left(\frac{4}{3}\right)}{\left(-\frac{16}{3}\right)}=\frac{25}{4}$
and $y_1=\frac{-b^2}{c}=\frac{-16}{-16 / 3}=3$
$\therefore$ the point of contact is $P\left(\frac{25}{4}, 3\right)$

Comparing this equation with $\frac{x^2}{ a ^2}-\frac{y^2}{ b ^2}=1$,

we get

$a ^2=16$ and $b ^2=9$

Let $P \left(x_1, 3\right)$ be the point on the hyperbola in the

first quadrant at which the tangent is drawn.

Substituting $x=x_1$ and $y=3$ in equation of

hyperbola, we get

$\begin{array}{ll} & \frac{x_1^2}{16}-\frac{3^2}{9}=1 \\ \therefore \quad & \frac{x_1^2}{16}-1=1 \\ \therefore \quad & \frac{x_1^2}{16}=2 \\ \therefore \quad & x_1^2=32 \\ \therefore \quad & x_1= \pm 4 \sqrt{2}\end{array}$

Since $P$ lies in the first quadrant,

$P \equiv(4 \sqrt{2}, 3)$

Equation of the tangent to the hyperbola

$\frac{x^2}{ a ^2}-\frac{y^2}{ b ^2}=1$ at $\left(x_1, y_1\right)$ is

$\frac{x x_1}{ a ^2}-\frac{y y_1}{ b ^2}=1$

Equation of the tangent at $P(4 \sqrt{2}, 3)$ is

$\begin{aligned} & \frac{4 \sqrt{2} x}{16}-\frac{3 y}{9}=1 \\ \therefore \quad & \frac{\sqrt{2}}{4} x-\frac{y}{3}=1 \\ \therefore \quad & 3 \sqrt{2} x-4 y=12\end{aligned}$

$\frac{x^2}{4}+\frac{y^2}{3}=1$

Comparing this equation with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we get

$a^2=4, b^2=3$

Equations of tangents to the ellipse $\frac{x^2}{\sigma^2}+\frac{y^2}{b^2}=1$ having slope $m$ are

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

Here, $m=-\frac{1}{2}$

Equations of the tangents are

$\begin{aligned} & y=\frac{-1}{2} x \pm \sqrt{4\left(-\frac{1}{2}\right)^2+3}=\frac{-1}{2} x \pm 2 \\ & 2 y=-x \pm 4 \\ & x+2 y \pm 4=0\end{aligned}$

Consider the tangent x + 2y – 4 = 0

Let this tangent intersect the $\mathrm{X}$-axis at $\mathrm{A}\left(\mathrm{x}_1, 0\right)$ and $\mathrm{Y}$-axis at $\mathrm{B}\left(0, \mathrm{y}_1\right)$.

$\begin{aligned} & x_1+0-4=0 \text { and } 0+2 y_1-4=0 \\ & x_1=4 \text { and } \mathrm{y}_1=2 \\ & \mathrm{~A}=(4,0) \text { and } B=(0,2) \\ & \mathrm{I}(\mathrm{OA})=4 \text { and } \mathrm{I}(\mathrm{OB})=2\end{aligned}$

Area of $\triangle \mathrm{AOB}=\frac{1}{2} \times \mathrm{I}(\mathrm{OA}) \times \mathrm{I}(\mathrm{OB})$

$\begin{aligned} & =\frac{1}{2} \times 4 \times 2 \\ & =4 \text { sq.units }\end{aligned}$

Given, eccentricity $(e)=\frac{2}{3}$

The ellipse passes through $\left(2, \frac{-5}{3}\right)$.

Substituting $x=2$ and $y=\frac{-5}{3}$ in equation of ellipse, we get

$\begin{aligned} & \frac{2^2}{a^2}+\frac{\left(-\frac{5}{3}\right)^2}{b^2}=1 \\ & \frac{4}{a^2}+\frac{25}{9 \mathrm{~b}^2}=1 \\ & \frac{4}{\mathrm{a}^2}+\frac{25}{9 \mathrm{a}^2\left(1-\mathrm{e}^2\right)}=1 \quad \ldots\left[\because \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)\right]\end{aligned}$

Multiplying throughout by $a^2$, we get

$4+\frac{25}{9\left(1-e^2\right)}=a^2$

$4+\frac{25}{9\left[1-\left(\frac{2}{3}\right)^2\right]}=a^2$

$4+\frac{25}{9\left(1-\frac{4}{9}\right)}=a^2$

$\begin{aligned} & 4+\frac{25}{5}=a^2 \\ & 4+5=a^2 \\ & a^2=9\end{aligned}$

Now, $\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)$

$\begin{aligned} & =9\left[1-\left(\frac{2}{3}\right)^2\right] \\ & =9\left(1-\frac{4}{9}\right)=9\left(\frac{5}{9}\right)=5\end{aligned}$

The required equation of ellipse is $\frac{x^2}{9}+\frac{y^2}{5}=1$.

Distance between directrices $=\frac{2 a}{e}$

Given, distance between directrices = 10

$\frac{2 a}{e}=10$

a = 5e …..(i)

The ellipse passes through (-√5, 2).

Substituting x = -√5 and y = 2 in equation of ellipse, we get

$\begin{aligned} & \frac{(-\sqrt{5})^2}{a^2}+\frac{2^2}{b^2}=1 \\ & \frac{5}{a^2}+\frac{4}{b^2}=1 \\ & \frac{5}{a^2}+\frac{4}{a^2\left(1-e^2\right)}=1 \quad \ldots\left[\because b^2=a^2\left(1-e^2\right)\right]\end{aligned}$

Multiplying throughout by $\mathrm{a}^2$, we get

$5+\frac{4}{1-e^2}=a^2$

$5+\frac{4}{1-\mathrm{e}^2}=(5 \mathrm{e})^2 \quad \ldots[$ From (i) $]$

$\begin{aligned} & 5\left(1-\mathrm{e}^2\right)+4=25 \mathrm{e}^2\left(1-\mathrm{e}^2\right) \\ & 5-5 \mathrm{e}^2+4=25 \mathrm{e}^2-25 \mathrm{e}^4 \\ & 25 \mathrm{e}^4-30 \mathrm{e}^2+9=0 \\ & \left(5 \mathrm{e}^2-3\right)^2=0 \\ & 5 \mathrm{e}^2-3=0 \\ & \mathrm{e}^2=\frac{3}{5}\end{aligned}$

From $(\mathrm{i}), \mathrm{a}=5 \mathrm{e}$

$\begin{aligned} & a^2=25 e^2=25 \times \frac{3}{5} \\ & a^2=15\end{aligned}$

We know that,

$\begin{aligned} & b^2=a^2\left(1-\mathrm{e}^2\right) \\ & \mathrm{b}^2=15\left(1-\frac{3}{5}\right)\end{aligned}$

$\begin{aligned} & b^2=15\left(\frac{2}{5}\right) \\ & b^2=6\end{aligned}$

The required equation of ellipse is $\frac{x^2}{15}+\frac{y^2}{6}=1$.

The ellipse passes through the points (-3, 1) and (2, -2).

Substituting x = -3 and y = 1 in equation of ellipse, we get

$\begin{aligned} & \frac{(-3)^2}{a^2}+\frac{1^2}{b^2}=1 \\ & \frac{9}{a^2}+\frac{1}{b^2}=1\end{aligned}$

$\ldots$..(i)

Substituting $x=2$ and $y=-2$ in equation of

ellipse, we get

$\frac{2^2}{a^2}+\frac{(-2)^2}{b^2}=1$

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

$\ldots$..(ii)

Let $\frac{1}{\mathrm{a}^2}=\mathrm{A}$ and $\frac{1}{\mathrm{~b}^2}=\mathrm{B}$

Equations (i) and (ii) become

9A + B = 1 …..(iii)

4A + 4B = 1 …..(iv)

Multiplying (iii) by 4, we get

36A + 4B = 4 …..(v)

Subtracting (iv) from (v), we get

32A = 3

$\mathrm{A}=\frac{3}{32}$

Substituting $A=\frac{3}{32}$ in (iv), we get

$\begin{aligned} & 4\left(\frac{3}{32}\right)+4 B=1 \\ & \frac{3}{8}+4 B=1 \\ & 4 B=1-\frac{3}{8} \\ & 4 B=\frac{5}{8} \\ & B=\frac{5}{32}\end{aligned}$

Since $\frac{1}{a^2}=A$ and $\frac{1}{b^2}=B$,

$\frac{1}{a^2}=\frac{3}{32}$ and $\frac{1}{b^2}=\frac{5}{32}$

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

The required equation of ellipse is

$\frac{x^2}{\left(\frac{32}{3}\right)}+\frac{y^2}{\left(\frac{32}{5}\right)}$, i.e., $3 x^2+5 y^2=32$

Let the required equation of ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a>b$.

Length of latus rectum $=\frac{2 b^2}{a}$

$\begin{aligned} & \frac{2 b^2}{a}=6 \\ & b^2=3 a \ldots .(i)\end{aligned}$

Co-ordinates of foci are (±ae, 0)

ae = 2

$\begin{aligned} & a^2 e^2=4 \ldots \text { (ii) } \\ & \text { Now, } b^2=a^2\left(1-e^2\right) \\ & b^2=a^2-a^2 e^2 \\ & 3 a=a^2-4 \ldots \ldots[\text { From (i) and (ii)] } \\ & a^2-3 a-4=0 \\ & a^2-4 a+a-4=0 \\ & a(a-4)+1(a-4)=0 \\ & (a-4)(a+1)=0 \\ & a-4=0 \text { or } a+1=0 \\ & a=4 \text { or } a=-1\end{aligned}$

Since a = -1 is not possible,

a = 4

$a^2=16$

$b^2=3(4)=12$

The required equation of ellipse is $\frac{x^2}{16}+\frac{y^2}{12}=1$.

Distance between foci = 2ae Given,

distance between foci = 6

2ae = 6

ae = 3

$a=\frac{3}{e} \ldots \ldots$ (i)

Distance between directrices $=\frac{2 \pi}{e}$

Given, distance between directrices $=\frac{50}{3}$

$\begin{aligned} & \frac{2 \mathrm{a}}{\mathrm{e}}=\frac{50}{3} \\ & \frac{\mathrm{a}}{\mathrm{e}}=\frac{25}{3} \\ & \frac{3}{\mathrm{e}}=\frac{25}{3}\end{aligned}$

...[From (i)]

$\begin{aligned} & \frac{3}{e^2}=\frac{25}{3} \\ & \mathrm{e}^2=\frac{9}{25} \\ & \mathrm{e}=\frac{3}{5}\end{aligned}$

$\ldots[\therefore 0<\mathrm{e}<1]$

Substituting e $=\frac{3}{5}$ in (i), we get

$\begin{aligned} & a=\frac{3}{\frac{3}{5}} \\ & a=5 \\ & a^2=25\end{aligned}$

Now, $\begin{aligned} \mathrm{b}^2 & =\mathrm{a}^2\left(1-\mathrm{e}^2\right) \\ & =25\left[1-\left(\frac{3}{5}\right)^2\right] \\ & =25\left(1-\frac{9}{25}\right)=25\left(\frac{16}{25}\right)=16\end{aligned}$

The required equation of ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$.

$\Rightarrow x^2=-16 y$

Comparing this equation with $\mathrm{x}^2=-4$ by, we get

⇒ 4b = 16

'⇒ b = 4

Focus = S(0, -b) = (0, -4)

Centre of the circle is C(4, -1) and it passes through focus S of the parabola.

Radius = CS

$\begin{aligned} & =\sqrt{(4-0)^2+(-1+4)^2} \\ & =\sqrt{16+9} \\ & =5\end{aligned}$

Equation of the directrix is y – b = 0, i.e.,y – 4 = 0

Length of the perpendicular from centre C(4, -1) to the directrix

$\begin{aligned} & =\left|\frac{0(4)+1(-1)-4}{\sqrt{(0)^2+(1)^2}}\right| \\ & =\left|\frac{-1-4}{1}\right| \\ & =5 \\ & =\text { radius }\end{aligned}$

∴ The circle touches the directrix of the parabola.

Let A be the centre of the bridge.

From the figure, vertex of parabola is at A(0, 5).

Let the equation of parabola be

$x^2=4 b(y-5) \ldots(i)$

Since the parabola passes through (100, 30).

Substituting x = 100 and y = 30 in (i), we get

$\begin{aligned} & \Rightarrow 100^2=4 b(30-5) \\ & \Rightarrow 100^2=4 b(25) \\ & \Rightarrow 100^2=100 b \\ & \Rightarrow b=100\end{aligned}$

Substituting the value of b in (i), we get

$x^2=400(y-5) \ldots($ (ii)

Let l metres be the length of vertical supporting cable.

Then P(30, l) lies on (ii).

$\begin{aligned} & \Rightarrow 30^2=400(1-5) \\ & \Rightarrow 900=400(1-5) \\ & \Rightarrow \frac{9}{4}=1-5 \\ & \Rightarrow 1=\frac{9}{4}+5 \\ & \Rightarrow 1=\frac{9}{4} m=7.25 \mathrm{~m}\end{aligned}$

The length of the vertical supporting cable is 7.25 m.

Comparing this equation with $y^2=4 a x$, we get

⇒ 4a = 4

⇒ a = 1

Let the equation of common tangent be

$y=m x+\frac{1}{m} \ldots(i)$

Substituting $y=m x+\frac{1}{m}$ in $x^2=32 y$, we get

$\begin{aligned} & \Rightarrow x^2=32\left(m x+\frac{1}{m}\right)=32 m x+\frac{32}{m} \\ & \Rightarrow m x^2=32 m^2 x+32 \\ & \Rightarrow m x^2-32 m^2 x-32=0 \ldots \ldots \text { (ii) }\end{aligned}$

Line (i) touches the parabola $x^2=32 y$.

The quadratic equation (ii) in x has equal roots.

Discriminant = 0

$\begin{aligned} & \Rightarrow\left(-32 m^2\right)^2-4(m)(-32)=0 \\ & \Rightarrow 1024 m^4+128 m=0 \\ & \Rightarrow 128 m\left(8 m^3+1\right)=0 \\ & \Rightarrow 8 m^3+1=0 \ldots \ldots[\because m \neq 0] \\ & \Rightarrow m^3=-\frac{1}{8} \\ & \Rightarrow m=-\frac{1}{2}\end{aligned}$

Substituting $m=-\frac{1}{2}$ in $(\mathrm{i})$, we get

$\begin{aligned} & \Rightarrow y=-\frac{1}{2} x+\frac{1}{\left(-\frac{1}{2}\right)} \\ & \Rightarrow y=-\frac{1}{2} x-2\end{aligned}$

⇒ x + 2y + 4 = 0, which is the equation of the common tangent.