MATHS 1 Marks Question

According to the Pythogoras theorem, we have:

$\text{OA}^2+\text{OB}^2=\text{AB}^2$

From the figure, we can see that

$\text{OA}=\sqrt{\big(\frac{\text{b}^2}{\text{a}}-0\big)^2+(\text{ae}-0)^2}=\sqrt{\frac{\text{b}^4}{\text{a}^2}}+\text{a}^2\text{e}^2=\text{OB}$

and $\text{AB}=\frac{2\text{b}^2}{\text{a}}$

Now, $2\Big[\text{a}^2\text{e}^2+\frac{\text{b}^4}{\text{a}^2}\Big]=\frac{4\text{b}^4}{\text{a}^2}$

$\Rightarrow\text{a}^2\text{e}^2+\frac{\text{b}^4}{\text{a}^2}=\frac{2\text{b}^4}{\text{a}^2}$

$\Rightarrow\text{a}^2\text{e}^2=-\frac{\text{b}^4}{\text{a}^2}+\frac{2\text{b}^4}{\text{a}^2}$

$\Rightarrow\text{a}^2\text{e}^2=\frac{\text{b}^4}{\text{a}^2}$

$\Rightarrow\text{e}^2=\frac{\text{b}^4}{\text{a}^4}$

$\Rightarrow\text{e}=\frac{\text{b}^2}{\text{a}^2}$

We know that $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$

$\text{e}=\sqrt{1-\text{e}}$

On squaring both sides, we get:

$\text{e}^2+\text{e}-1=0$

$\Rightarrow\text{e}=\frac{-1\pm\sqrt{1+4}}{2}$ $\big(\because\ $Ecentricity cannot be negative $\big)$

$\Rightarrow\text{e}=\frac{\sqrt{5}-1}{2}$

We know that the focal distance of a point B(0, b) is $\text{a}\pm\text{e}.\ 0=\text{a}$

i. e. $\text{SB}=\text{SB}'=\text{a}$ 

$\therefore\text{SB}=\text{SB}'=2\text{a}$

since $\triangle\text{BSS}'$ is equilateral, we have:

$\text{SB}=\text{SS}'=\text{S}'\text{B}=2\text{ae}$

$\Rightarrow2\text{ae}+2\text{ae}=2\text{a}$

$\Rightarrow4\text{ae}=\text{a}$

$\Rightarrow\text{e}=\frac{2}{4}$

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

According to the question, the minor axis of the ellipse subtends an equilatreal triangle with vertex at one end of the major axis

$\text{}AB=\sqrt{\text{a}^2+\text{b}^2}$

We know that ABC is an equilateral triangle.

$\therefore\text{AB}=\text{BB}'$

$\Rightarrow\sqrt{\text{a}^2+\text{b}^2}=2\text{b}$

On squaring both sides, we have:

$\Rightarrow{\text{a}^2+\text{b}^2}=4\text{b}$

$\Rightarrow\text{a}^2=3\text{b}^2$

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

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

$\Rightarrow\text{e}=\sqrt{1-\frac{1}{3}}$

$\Rightarrow\text{e}=\sqrt{\frac{2}{3}}$

The give equation of the ellipse is

$4\text{x}^2+9\text{y}^2=36\ \dots(\text{i})$

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

This is of the form $\Rightarrow\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1,$ where a² = 9 and b² = 4 i.e a = 3 and b = 2 clearly a > b, therefore the major axis and minir axis of the ellipse are along x axis and y axis respectively.

Let, e be the ecentricity of the ellipse. Then,

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

$\Rightarrow\text{e}=\sqrt{1-\frac{4}{9}}$

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

Therefore, coordinates of focus at S i.e $(\text{ae}, 0)=\big(\sqrt{5},0\big)$

& coordinates of foucs at S'  $=(-\text{ae},0)=\big(-\sqrt{5},0\big)$

It is given that PSQ is a focal chord

As we know that,

$\text{SP}+\text{S}'\text{P}=2\text{a}$

$\Rightarrow4+\text{S}'\text{P}=6$ $\big[\because\ $SP = 4 $\big]$

$\Rightarrow\text{S}'\text{P}=2$

Let coordinates of P be (m, n)

As $\text{S}'\text{P}=2$

$\therefore\sqrt{(\text{n}-0)^2+(\text{m}+\sqrt{5})^2}=2$

$\Rightarrow\text{n}^2+\text{m}^2+5+2\sqrt{5}\text{m}=4\ \dots(\text{ii})$

& SP = 4

$\therefore\sqrt{(\text{n}-0)^2+(\text{m}+\sqrt{5})^2}=4$

$\Rightarrow\text{n}^2+\text{m}^2+5+2\sqrt{5}\text{m}=16\ \dots(\text{iii})$

Subtracting eq (iii) from eq (ii), we get

$4\sqrt{5}\text{m}=-12$

$\Rightarrow\text{m}=\frac{-3}{\sqrt{5}}$

& we get $\text{n}=\frac{4}{\sqrt{5}}$

$\therefore\ $coordinates of a P are $\Big(\frac{-3}{\sqrt{5}},\frac{4}{\sqrt{5}}\Big)$ and coordinates of S are $\big(\sqrt{5},0\big)$

$\therefore\ $Equation of the line segementf PS which is extended to PQ is given by 

$\text{x}+2\text{y}=\sqrt{5}\ \dots(\text{iv})$

solving eq (i) & (iv) we get,

$\text{x}=\frac{-3}{\sqrt{5}}$ and $\text{y}=\frac{4}{\sqrt{5}}$ which would be the coordinates of Q.

$\therefore\text{S}'Q=\sqrt{\Big(\frac{-8\sqrt{5}}{50}-0\Big)^2+\Big(\frac{66\sqrt{5}}{50}+\sqrt{5}\Big)^2}$

$=\frac{26}{5}$