Find the length of the line-segment joining

Solution:

We know that equation of an ellipse is $y^{2}=4 a x$, Also we have Length of latus rectum $=4 a$

Let the point on parabola be (i, j)

From the figure, slope of $O P=\tan \theta=\frac{j}{i}$

$\mathrm{j}=i \tan \theta$

Squaring the above equation on both sides we get

$j^{2}=i^{2} \tan ^{2} \theta$

$j^{2}=4 a i$

$O P=\sqrt{i^{2}+j^{2}}$

By substituting the value of $j$ then we get

Squaring the above equation on both sides we get

$\mathrm{j}^{2}=\mathrm{i}^{2} \tan ^{2} \theta$

$j^{2}=4 a i \ldots \ldots .2$

$O P=\sqrt{\dot{1}^{2}+j^{2}}$

By substituting the value of $j$ then we get

$O P=\sqrt{i^{2}+(i \tan \theta)^{2}}$

$=\sqrt{i^{2}+i^{2} \tan ^{2} \theta}$

Taking $i^{2}$ common we get

$=\sqrt{i^{2}\left(1+\tan ^{2} \theta\right)}$

We know that

$1+\tan ^{2} \theta=\operatorname{Sec}^{2} \theta$

$=\sqrt{{i}^{2} \operatorname{Sec}^{2} \theta}$

The above equation can be written as

$=1 \operatorname{Sec} \theta$

Now by equating the equations 1 and 2 we get

$4 a i=i^{2} \tan ^{2} \theta$

On rearranging we get

$\mathrm{i}^{2} \tan ^{2} \theta-4 \mathrm{ai}=0$

Taking I common

$\mathrm{i}\left(\mathrm{i} \tan ^{2} \theta-4 \mathrm{a}\right)=0$

$\mathrm{i}=0$ which is not Possible

Or

$\operatorname{itan}^{2} \theta-4 a=0$

$\mathrm{i}=\frac{4 \mathrm{a}}{\tan ^{2} \theta}$\

$j=\frac{4 a}{\tan ^{2} \theta}$

Now $\mathrm{OP}=\mathrm{i} \operatorname{Sec} \theta=\frac{4 \mathrm{a}}{\tan ^{2} \theta} \times \operatorname{Sec} \theta=\frac{4 \mathrm{ax} \operatorname{Cos}^{2} \theta \times 1}{\operatorname{Sin}^{2} \theta \times \operatorname{Cos} \theta}=4 \mathrm{a} \operatorname{Cot} \theta \operatorname{Cosec} \theta$

Hence, the length of line segment is $4 \mathrm{aCot} \theta \operatorname{Cosec} \theta$ units.