Maths Solve the Following Question.(2 Marks)

$\therefore \frac{x^2}{16}-\frac{y^2}{48}=1$

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

$a^2=16$ and $b^2=48$

Slope of the line $x+2 y-7=0$ is $-\frac{1}{2}$

Since the given line is perpendicular to the tangents,

slope of the required tangent (m) = 2

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

Equations of tangents to the ellipse having slope m are

$\begin{aligned} & y=m x \pm \sqrt{a^2 m^2-b^2} \\ & y=2 x \pm \sqrt{16(2)^2-48} \\ & y=2 x \pm \sqrt{16} \\ & \therefore y=2 x \pm 4\end{aligned}$

$\therefore \frac{x^2}{100}+\frac{y^2}{25}=1$

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

$a^2=100$ and $b^2=25$

Equation of tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $\left( x _1, y _1\right)$ is $\frac{x x_1}{a^2}+\frac{y y_1}{b^2}=1$

Equation of tangent at (8, 3) is

$\begin{aligned} & \frac{8 x}{100}+\frac{3 y}{25}=1 \\ & \frac{2 x}{25}+\frac{3 y}{25}=1 \\ & 2 x+3 y=25\end{aligned}$

$m=-\frac{2}{3}$ and $c=4$
We know that,
if the line $y=m x+c$ is tangent to the ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ then $c^2=a^2 m^2+b^2$.
Here $c^2=(4)^2=16$ and
$
\begin{aligned}
& a^2 m^2+b^2=(18)\left(-\frac{2}{3}\right)^2+(8)=(18)\left(\frac{4}{9}\right)+8 \\
& =(2)(4)+8=16
\end{aligned}
$
hence the given line is tangent to the given ellipse.

$
\frac{x}{4}+\frac{\sqrt{3} y}{6}=1
$
i.e. $6 x+4 \sqrt{3} y=24$
i.e. $3 x+2 \sqrt{3} y=12$
Thus required equation of tangent is
$
3 x+2 \sqrt{3} y=12 \text {. }
$

$
\begin{aligned}
& \therefore-9 m=-4 m^2+4 \\
& \therefore 4 m^2-9 m+4=0
\end{aligned}
$
$m_1$ and $m_2$ be the slopes (roots)
$
\begin{aligned}
& \left(m_1-m_2\right)=+\frac{\text { constant }}{\text { co-efficient of } m^2} \\
& m_1 \cdot m_2=-\frac{4}{4} \quad \therefore m_1 \cdot m_2=-1
\end{aligned}
$
hence tangents are perpendicular to each other.

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

Distance between foci = 2ae

According to the given condition,

distance between directrices = 3(distance between foci)

$\begin{aligned} & \frac{2 a}{e}=3(2 a e) \\ & \frac{1}{e}=3 e \\ & \frac{1}{3}=e^2 \\ & e=\frac{1}{\sqrt{3}} \ldots \ldots[\because 0<e<1]\end{aligned}$

Eccentricity of the ellipse is $\frac{1}{\sqrt{3}}$

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

⇒ 4a = 4

⇒ a = 1

Focal distance of a point = x + a

Given, focal distance = 17

⇒ x + 1 = 17

⇒ x = 16

Substituting $x=16$ in $y^2=4 x_r$ we get

$\begin{aligned} & \Rightarrow y^2=4(16) \\ & \Rightarrow y^2=64 \\ & \Rightarrow y= \pm 8\end{aligned}$

∴ The co-ordinates of the point on the parabola are (16, 8) or (16, -8).

$\Rightarrow y^2=\frac{7}{2} x$

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

$\begin{aligned} & \Rightarrow 4 a=\frac{7}{2} \\ & \Rightarrow a=\frac{7}{8}\end{aligned}$

If t is the parameter of the point P on the parabola, then

$\begin{aligned} & P(t)=\left(a t^2, 2 a t\right) \\ & \text { i.e., } x=a t^2 \text { and } y=2 a t\end{aligned}$

…..(i)

Given, t = -2

Substituting $a=\frac{7}{8}$ and $t=-2$ in (i), we get

$\begin{aligned} & x=\frac{7}{8}(-2) 2 \text { and } y=2\left(\frac{7}{8}\right)(-2) \\ & x=\frac{7}{2} \text { and } y=\frac{-7}{2}\end{aligned}$

The co-ordinates of the point on the parabola are $\left(\frac{7}{2}, \frac{-7}{2}\right)$

∴ Focal distance = x + a

$\begin{aligned} & =\frac{7}{2}+\frac{7}{8} \\ & =\frac{35}{8}\end{aligned}$

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

⇒ 4a = 12

⇒ a = 3

If t is the parameter of the point P on the parabola, then

$\begin{aligned} & P(t)=\left(a t^2, 2 a t\right) \\ & \text { i.e., } x=a t^2 \text { and } y=2 a t\end{aligned}$

Given, $\mathrm{t}=\frac{1}{3}$

Substituting $a=3$ and $t=\frac{1}{3}$ in (i), we get

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

$x=\frac{1}{3}$ and $y=2$

The co-ordinates of the point on the parabola are $\left(\frac{1}{3}, 2\right)$

∴ Focal distance = x + a

$\begin{aligned} & =\frac{1}{3}+3 \\ & =\frac{10}{3}\end{aligned}$

Comparing this equation with $y^2=4 a x_{\text {, }}$ we get

⇒ 4a = 16

⇒ a = 4

Since ordinate is 2 times the abscissa,

y = 2x

Substituting $y=2 x$ in $y^2=16 x$, we get

$\begin{aligned} & \Rightarrow(2 x)^2=16 x \\ & \Rightarrow 4 x^2=16 x \\ & \Rightarrow 4 x^2-16 x=0 \\ & \Rightarrow 4 x(x-4)=0 \\ & \Rightarrow x=0 \text { or } x=4\end{aligned}$

When x = 4,

focal distance = x + a = 4 + 4 = 8

When x = 0,

focal distance = a = 4

∴ Focal distance is 4 or 8.

Substituting $x=27, y=-12$ and $a=\frac{4}{3}$ in (i), we get

$\begin{aligned} & 27=\frac{4}{3} t^2 \text { and }-12=2\left(\frac{4}{3}\right) t \\ & t^2=\frac{81}{4} \text { and } t=\frac{-9}{2} \\ & t= \pm \frac{9}{2} \text { and } t=\frac{-9}{2}\end{aligned}$

$t=\frac{-9}{2}$

$\therefore$ The parameter of the given point is $\frac{-9}{2}$

$\Rightarrow y^2=\frac{16}{3} x$

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

$\begin{aligned} & \Rightarrow 4 a=\frac{16}{3} \\ & \Rightarrow a=\frac{4}{3}\end{aligned}$

If t is the parameter of the point P on the parabola, then

$\begin{aligned} & P(t)=\left(a t^2, 2 a t\right) \\ & \text { i.e., } x=a t^2 \text { and } y=2 a t\end{aligned}$

………(i)

(i) Given point is (3, -4)

Substituting $x=3, y=-4$ and $a=\frac{4}{3}$ in (i), we get

$\begin{aligned} & 3=\frac{4}{3} t^2 \text { and }-4=2\left(\frac{4}{3}\right) t \\ & t^2=\frac{9}{4} \text { and } t=\frac{-3}{2} \\ & t= \pm \frac{3}{2} \text { and } t=\frac{-3}{2} \\ & t=-\frac{3}{2}\end{aligned}$

$\therefore$ The parameter of the given point is $\frac{-3}{2}$

Equation of the parabola can be either $y^2=4 a x$ or $y^2=-4 a x$.

Since the parabola passes through (2, 3), it lies in 1st quadrant.

$\therefore$ Required parabola is $y^2=4 a x$.

Substituting $x=2$ and $y=3$ in $y^2=4 a x$, we get

$\begin{aligned} & \Rightarrow(3)^2=4 a(2) \\ & \Rightarrow 9=8 a \\ & \Rightarrow a=\frac{9}{8}\end{aligned}$

The required equation of the parabola is

$\begin{aligned} & y^2=4\left(\frac{9}{8}\right) x \\ & \Rightarrow y^2=\frac{9}{2} x \\ & \Rightarrow 2 y^2=9 x\end{aligned}$

Equation of the parabola can be either $y^2=4 a x$ or $y^2=-4 a x$.

Since the parabola passes through (1, -6), it lies in the 4th quadrant.

Required parabola is $y^2=4 a x$.

Substituting $x=1$ and $y=-6$ in $y^2=4 a x$, we get

$\begin{aligned} & \Rightarrow(-6)^2=4 a(1) \\ & \Rightarrow 36=4 a \\ & \Rightarrow a=9\end{aligned}$

$\therefore$ The required equation of the parabola is $y^2=4(9) x$, i.e. $y^2=36 x$.

Since focus lies on X-axis, it is the axis of the parabola.

Focus S(-7, 0) lies on the left-hand side of the origin.

It is a left-handed parabola.

Required parabola is y = -4ax.

Focus is S(-a, 0).

a = 7

$\therefore$ The required equation of the parabola is $y^2=-4(7) x$, i.e, $y^2=-28 x$.

Since the parabola passes through (3, 4), it lies in the 1st quadrant.

Required parabola is $y^2=4 a x$.

Substituting $x=3$ and $y=4$ in $y^2=4 a x$, we get

$\begin{aligned} & \Rightarrow(4)^2=4 a(3) \\ & \Rightarrow a=\frac{16}{12}=\frac{4}{3}\end{aligned}$

The required equation of the parabola is

$\begin{aligned} & y^2=4\left(\frac{4}{3}\right) x \\ & \Rightarrow 3 y^2=16 x\end{aligned}$

Equation of the parabola can be either $x^2=4$ by or $x^2=-4$ by

Since the parabola passes through (-10, -5), it lies in 3rd quadrant.

Required parabola is $\mathrm{x}^2=-4$ by.

Substituting $x=-10$ and $y=-5$ in $x^2=-4$ by, we get

$\begin{aligned} & \Rightarrow(-10)^2=-4 b(-5) \\ & \Rightarrow b=\frac{100}{20}=5\end{aligned}$

$\therefore$ The required equation of the parabola is $x^2=-4(5) y$, i.e., $x^2=-20 y$.

$\Rightarrow y^2=-\frac{16}{3} x$

Comparing this equation withy $=-4 a x$, we get

$\begin{aligned} & \Rightarrow 4 a=\frac{16}{3} \\ & \Rightarrow a=\frac{4}{3}\end{aligned}$

Co-ordinates of focus are $S(-a, 0)$, i.e, $\left(-\frac{4}{3}, 0\right)$

Equation of the directrix is x – a = 0,

$\begin{aligned} & \Rightarrow x--\frac{4}{3}=0 \\ & \Rightarrow 3 x-4=0\end{aligned}$

Length of latus rectum $=4 a=4\left(\frac{4}{3}\right)=\frac{16}{3}$

Co-ordinates of end points of latus rectum are (-a, 2a) and (-a, -2a),

i.e., $\left(-\frac{4}{3}, \frac{8}{3}\right)$ and $\left(-\frac{4}{3},-\frac{8}{3}\right)$

Comparing this equation with $x^2=-4 b y_{\text {, }}$ we get

⇒ 4b = 8

⇒ b = 2

Co-ordinates of focus are S(0, -b), i.e., S(0, – 2)

Equation of the directrix is y – b = 0, i.e., y – 2 = 0

Length of latus rectum = 4b = 4(2) = 8

∴ Co-ordinates of end points of latus rectum are (2b, -b) and (-2b, -b), i.e., (4, -2) and (-4, -2).

$\Rightarrow x^2=\frac{8}{3} y$

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

$\begin{aligned} & \Rightarrow 4 b=\frac{8}{3} \\ & \Rightarrow b=\frac{2}{3}\end{aligned}$

Co-ordinates of focus are $S\left(0\right.$, b), i.e., $S\left(0, \frac{2}{3}\right)$

Equation of the directrix is y + b = 0,

$\Rightarrow y+\frac{2}{3}=0$

⇒ 3y + 2 = 0

Length of latus rectum $=4 \mathrm{~b}=4\left(\frac{2}{3}\right)=\frac{8}{3}$

Co-ordinates of end points of latus rectum are $(2 b, b)$ and $(-2 b, b)$,

$\Rightarrow\left(\frac{4}{3}, \frac{2}{3}\right)$ and $\left(-\frac{4}{3}, \frac{2}{3}\right)$.

Comparing this equation with $y^2=-4 a x_{\text {}}$ we get

⇒ 4a = 20

⇒ a = 5

Co-ordinates of focus are S(-a, 0), i.e., S(-5, 0)

Equation of the directrix is x – a = 0

⇒ x – 5 = 0

Length of latus rectum = 4a = 4(5) = 20

Co-ordinates of end points of latus rectum are (-a, 2a) and (-a, -2a),

⇒ (-5, 10) and (-5, -10).

$\Rightarrow y^2=\frac{24}{5} x$

Comparing this equation with $y^2=4 a x_{\text {}}$ we get

$\begin{aligned} & \Rightarrow 4 a=\frac{24}{5} \\ & \Rightarrow a=\frac{6}{5}\end{aligned}$

Co-ordinates of focus are $S(a, 0)$, i.e., $S\left(\frac{6}{5}, 0\right)$

Equation of the directrix is x + a = 0.

$\begin{aligned} & \Rightarrow x+\frac{6}{5}=0 \\ & \Rightarrow 5 x+6=0\end{aligned}$

Length of latus rectum = 4a

$\begin{aligned} & =4\left(\frac{6}{5}\right) \\ & =\frac{24}{5}\end{aligned}$

Co-ordinates of end points of latus rectum are (a, 2a) and (a, -2a)

$\Rightarrow\left(\frac{6}{5}, \frac{12}{5}\right)$ and $\left(\frac{6}{5}, \frac{-12}{5}\right)$