Maths Solve the Following Question.(5 Marks)

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

$\begin{aligned} & \therefore a^2=25, b^2=16 \\ & \therefore a=5, b=4\end{aligned}$

We know that $e =\frac{\sqrt{a^2-b^2}}{a}$

$\begin{aligned} & \therefore e=\frac{\sqrt{25-16}}{5}=\frac{3}{5} \\ & a e=5\left(\frac{3}{5}\right)=3\end{aligned}$

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

i.e., S(3, 0) and S'(-3, 0)

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

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

Equation of one of the tangents to the ellipse is

$\begin{aligned} & y=m x+\sqrt{25 m^2+16} \\ & \therefore m x-y+\sqrt{25 m^2+16}=0\end{aligned}$

p1 = length of perpendicular segment from S(3, 0) to the tangent (i)

$\begin{aligned} & =\left|\frac{ m (3)-0+\sqrt{25 m ^2+16}}{\sqrt{ m ^2+1}}\right| \\ & \therefore p _1=\left|\frac{3 m +\sqrt{25 m ^2+16}}{\sqrt{ m ^2+1}}\right|\end{aligned}$

p2 = length of perpendicular segment from S'(-3, 0) to the tangent (i)

$\begin{aligned} & =\left|\frac{ m (-3)-0+\sqrt{25 m^2+16}}{\sqrt{m^2+1}}\right| \\ & \therefore p_2=\left|\frac{-3 m+\sqrt{25 m^2+16}}{\sqrt{m^2+1}}\right|\end{aligned}$

$\begin{aligned} & \therefore p_1 p_2=\left|\frac{3 m+\sqrt{25 m^2+16}}{\sqrt{m^2+1}}\right|\left|\frac{-3 m+\sqrt{25 m^2+16}}{\sqrt{m^2+1}}\right| \\ & =\frac{\left(25 m^2+16\right)-9 m^2}{m^2+1} \\ & =\frac{16\left(m^2+1\right)}{m^2+1} \\ & =16\end{aligned}$

Parametric form x = a cosθ , y = b sinθ That is x = 4 cosθ , y = 3 sinθ

Comparing this equation with $\frac{x^2}{3^2}+\frac{y^2}{h^2}=1$, we get

$a^2=25, b^2=16$

a = 5, b = 4

We know that $\mathrm{e}=\frac{\sqrt{\mathrm{a}^2-\mathrm{b}^2}}{\mathrm{a}}$

$e=\frac{\sqrt{25-16}}{5}=\frac{\sqrt{9}}{5}=\frac{3}{5}$

ae $=5\left(\frac{3}{5}\right)=3$

Co-ordinates of foci are $S(a e, 0)$ and $S^{\prime}(-\mathrm{ae}, 0)$,

i.e. $S(3,0)$ and $S^{\prime}(-3,0)$

Equations of tangents to the ellipse

$\frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1$ having slope $\mathrm{m}$ are

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

Equation of one of the tangents to the ellipse is

$\begin{aligned} & y=m x+\sqrt{25 m^2+16} \\ & m x-y+\sqrt{25 m^2+16}=0\end{aligned}$

$\ldots$...(i)

$\mathrm{p}_1=$ length of perpendicular segment from

$S(3,0)$ to the tangent (i)

$\begin{aligned} & =\left|\frac{m(3)-0+\sqrt{25 m^2+16}}{\sqrt{m^2+1}}\right| \\ & p_1=\left|\frac{3 m+\sqrt{25 m^2+16}}{\sqrt{m^2+1}}\right|\end{aligned}$

$\mathbf{p}_2=$ length of perpendicular segment from

$S^{\prime}(-3,0)$ to the tangent (i)

$=\left|\frac{m(-3)-0+\sqrt{25 m^2+16}}{\sqrt{m^2+1}}\right|$

$p_2=\frac{-3 m+\sqrt{25 m^2}+16}{\sqrt{m^2+1}}$

$\begin{aligned} p_1 p_2= & \left|\frac{3 m+\sqrt{25 m^2+16}}{\sqrt{m^2+1}}\right|\left|\frac{-3 m+\sqrt{25 m^2+16}}{\sqrt{m^2+1}}\right| \\ & =\frac{\left(25 m^2+16\right)-9 m^2}{m^2+1}=\frac{16\left(m^2+1\right)}{m^2+1}=16\end{aligned}$

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

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

$\begin{aligned} & a^2=\frac{1}{3} \text { and } b^2=\frac{1}{4} \\ & a=\frac{1}{\sqrt{3}} \text { and } b=\frac{1}{2}\end{aligned}$

Since a > b,

X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis $=2 a=2\left(\frac{1}{\sqrt{3}}\right)=\frac{2}{\sqrt{3}}$

Length of minor axis $=2 b=2\left(\frac{1}{2}\right)=1$

Lengths of the principal axes are $\frac{2}{\sqrt{3}}$ and 1 .

(ii) We know that $\mathrm{e}=\frac{\sqrt{a^2-b^2}}{a}$

$e=\frac{\sqrt{\frac{1}{3}-\frac{1}{4}}}{\frac{1}{\sqrt{3}}}=\frac{\sqrt{\frac{1}{12}}}{\frac{1}{\sqrt{3}}}=\sqrt{\frac{3}{12}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

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

$\begin{aligned} & \text { i.e. } S\left(\frac{1}{\sqrt{3}}\left(\frac{1}{2}\right), 0\right) \text { and } S^{\prime}\left(-\frac{1}{\sqrt{3}}\left(\frac{1}{2}\right), 0\right) \\ & \text { i.e., } S\left(\frac{1}{2 \sqrt{3}}, 0\right) \text { and } S^{\prime}\left(-\frac{1}{2 \sqrt{3}}, 0\right)\end{aligned}$

(iii) Equations of the directrices are $x= \pm \frac{a}{e}$,

$\begin{aligned} & = \pm \frac{\frac{1}{\sqrt{3}}}{\frac{1}{2}} \\ & = \pm \frac{2}{\sqrt{3}}\end{aligned}$

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

$\begin{aligned} & =\frac{2\left(\frac{1}{2}\right)^2}{\frac{1}{\sqrt{3}}} \\ & =\frac{\sqrt{3}}{2}\end{aligned}$

$\begin{aligned} & =2\left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{2}\right) \\ & =\frac{1}{\sqrt{3}}\end{aligned}$

(v) Distance between foci = 2ae

$\begin{aligned} & =2\left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{2}\right) \\ & =\frac{1}{\sqrt{3}}\end{aligned}$

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

$\begin{aligned} & =\frac{2\left(\frac{1}{\sqrt{3}}\right)}{\frac{1}{2}} \\ & =\frac{4}{\sqrt{3}}\end{aligned}$

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

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

$\begin{aligned} & a^2=3 \text { and } b^2=1 \\ & a=\sqrt{3} \text { and } b=1\end{aligned}$

Since a > b,

X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2√3

Length of minor axis = 2b = 2(1) = 2

Lengths of the principal axes are 2√3 and 2.

(ii) We know that e $=\frac{\sqrt{a^2-b^2}}{a}$

$\begin{aligned} & =\frac{\sqrt{3-1}}{\sqrt{3}} \\ & =\frac{\sqrt{2}}{\sqrt{3}}\end{aligned}$

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

i.e., $S\left(\sqrt{ } 3\left(\frac{\sqrt{2}}{\sqrt{3}}\right), 0\right)$ and $S^{\prime}\left(-\sqrt{ } 3\left(\frac{\sqrt{2}}{\sqrt{3}}\right), 0\right)$

i.e., S(√2, 0) and S'(-√2, 0)

(iii) Equations of the directrices are $x= \pm \frac{a}{e}$,

$\begin{aligned} & = \pm \frac{\sqrt{3}}{\frac{\sqrt{2}}{\sqrt{3}}} \\ & = \pm \frac{3}{\sqrt{2}}\end{aligned}$

(iv) Length of latus rectum $=\frac{2 b^2}{a}=\frac{2(1)^2}{\sqrt{3}}=\frac{2}{\sqrt{3}}$

(v) Distance between foci = 2ae

$\begin{aligned} & =2(\sqrt{3})\left(\frac{\sqrt{2}}{\sqrt{3}}\right) \\ & =2 \sqrt{2}\end{aligned}$

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

$\begin{aligned} & =\frac{2 \sqrt{3}}{\frac{\sqrt{2}}{\sqrt{3}}} \\ & =\frac{2 \times 3}{\sqrt{2}} \\ & =3 \sqrt{2}\end{aligned}$

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

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

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

a = 2 and b = √3

Since a > b,

X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(2) = 4

Length of minor axis = 2b = 2√3

Lengths of the principal axes are 4 and 2√3.

(ii) We know that $\mathrm{e}=\frac{\sqrt{a^2-b^2}}{a}$

$\begin{aligned} & =\frac{\sqrt{4-3}}{2} \\ & =\frac{1}{2}\end{aligned}$

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

i.e., $S\left(2\left(\frac{1}{2}\right), 0\right)$ and $S^{\prime}\left(-2\left(\frac{1}{2}\right), 0\right)$

i.e., S(1, 0) and S'(-1, 0)

(iii) Equations of the directrices are $x= \pm \frac{\mathrm{a}}{\mathrm{e}}$

$\begin{aligned} & = \pm \frac{2}{\frac{1}{2}} \\ & = \pm 4\end{aligned}$

(iv) Length of latus rectum $=\frac{2 b^2}{a}=\frac{2(\sqrt{3})^2}{2}=3$

(v) Distance between foci $=2 a e=2(2)\left(\frac{1}{2}\right)=2$

$\begin{aligned} & =\frac{2(2)}{\frac{1}{2}} \\ & =8\end{aligned}$

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

$\begin{aligned} & a^2=25 \text { and } b^2=9 \\ & a=5 \text { and } b=3\end{aligned}$

Since a > b, X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(5) = 10 Length of minor axis = 2b = 2(3) = 6 Lengths of the principal axes are 10 and 6.

(ii) We know that $\mathrm{e}=\frac{\sqrt{a^2-b^2}}{a}$

$\begin{aligned} & =\frac{\sqrt{25-9}}{5} \\ & =\frac{\sqrt{16}}{5} \\ & =\frac{4}{5}\end{aligned}$

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

i.e., $S\left(5\left(\frac{4}{5}\right), 0\right)$ and $S^{\prime}\left(-5\left(\frac{4}{5}\right), 0\right)$

i.e., S(4, 0) and S'(-4, 0)

(iii) Equations of the directrices are $x= \pm \frac{a}{e}$

$\begin{aligned} & = \pm \frac{5}{\frac{4}{5}} \\ & = \pm \frac{25}{4}\end{aligned}$

(iv) Length of latus rectum $=\frac{2 b^2}{a}=\frac{2(3)^2}{5}=\frac{18}{5}$

(v) Distance between foci = 2ae

$\begin{aligned} & =2(5)\left(\frac{4}{5}\right) \\ & =8\end{aligned}$

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

$\begin{aligned} & =\frac{2(5)}{\frac{4}{5}} \\ & =\frac{25}{2}\end{aligned}$

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

⇒ 4a = 8 ⇒ a = 2

Equation of tangent to given parabola at $A\left(t_1\right)$ is $y$

$t_1=x+2 t_1^2 \ldots \ldots$ (i)

Equation of tangent to given parabola at $B\left(t_2\right)$ is $y$

$t_2=x+2 t_2^2 \ldots$ (ii)

A tangent at the vertex is Y-axis whose equation is x = 0.

x-coordinate of points P and Q is 0.

Let P be(0, k1) and Q be (0, k2).

Then, from (i) and (ii), we get

$\begin{aligned} & \mathrm{k}_1 \mathrm{t}_1=0+2 \mathrm{t}_1{ }^2 \text { and } \mathrm{k}_2 \mathrm{t}_2=0+2 \mathrm{t}_2{ }^2 \\ & \mathrm{k}_1=2 \mathrm{t}_1 \text { and } \mathrm{k}_2=2 t_2 \\ & \mathrm{P} \text { is }\left(0,2 \mathrm{t}_1\right) \text { and } \mathrm{Q} \text { is }\left(0,2 \mathrm{t}_2\right) \\ & \mathrm{PQ}=2\left|\mathrm{t}_1-\mathrm{t}_2\right|\end{aligned}$

But $P Q$ is given to be 4 .

$\begin{aligned} & 2\left|t_1-t_2\right|=4 \\ & \left(t_1-t_2\right)^2=4\end{aligned}$

...(iii)

Let $\mathrm{R}\left(x_1, y_1\right)$ be point of intersection of (i) and (ii).

$\begin{aligned} & y_1 t_1=x_1+2 t_1^2 \\ & \text { and } y_1 t_2=x_1+2 t_2^2\end{aligned}$

$\ldots$ (iv)

Subtracting, we get

$\begin{aligned} & y_1 t_1-y_1 t_2=2 t_1^2-2 t_2^2 \\ & y_1\left(t_1-t_2\right)=2\left(t_1+t_2\right)\left(t_1-t_2\right) \\ & y_1=2\left(t_1+t_2\right) \quad \ldots(v)\left[\because t_1 \neq t_2\right]\end{aligned}$

$\ldots(v)\left[\because t_1 \neq t_2\right]$

$\begin{aligned} & y_1 t_1=2\left(\mathrm{t}_1+\mathrm{t}_2\right) \mathrm{t}_1 \\ & x_1+2 t_1^2=2 t_1^2+2 t_1 t_2 \\ & \text {... From (iv)] } \\ & x_1=2 \mathrm{t}_1 \mathrm{t}_2 \\ & \end{aligned}$

To find the equation of locus of $\mathrm{R}\left(x_1, y_1\right)$,

eliminate $t_1$ and $t_2$ from the equations (iii), (v)

and (vi).

Squaring (v), we get

$\begin{aligned} y_1^2 & =4\left(\mathrm{t}_1+\mathrm{t}_2\right)^2 \\ & =4\left[\left(\mathrm{t}_1-\mathrm{t}_2\right)^2+4 \mathrm{t}_1 \mathrm{t}_2\right] \\ & =4\left[4+2 x_1\right] \quad \ldots[\text { From (iii) and (vi) }] . \\ y_1^2 & =8\left(x_1+2\right)\end{aligned}$

$\therefore$ Equation of locus of $R$ is $y^2=8(x+2)$.