Question

Prove that the extremities of latus recta of all ellipses a given major axis lie in the parabola ${x^2} = - a\left( {y - a} \right)$

Answer 1

Here we will use a known fact that for a standard ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$, $a > b$ the endpoints of latus recta are given by $\left( {ae, \pm \dfrac{{{b^2}}}{a}} \right)$
where $x = ae$ and $y = \pm \dfrac{{{b^2}}}{a}$

Complete step-by-step answer:
We know that for a standard ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$, $a > b$
The coordinates of endpoints of the latus rectum are:

$$x = ae \\\ \Rightarrow e = \dfrac{x}{a}........................\left( 1 \right) \\\$$

And,
$y = \dfrac{{{b^2}}}{a}$……………………….(2)
Now in order to prove that major axis lie in the parabola ${x^2} = - a\left( {y - a} \right)$
Now we also know that :-
${b^2} = {a^2}\left( {1 - {e^2}} \right)$………………….(3)
Putting the value of ${b^2}$in equation 2 we get:-

$$y = \dfrac{{{a^2}\left( {1 - {e^2}} \right)}}{a} \\\ \Rightarrow y = a\left( {1 - {e^2}} \right) \\\$$

Now putting the value of e from equation 1 we get:-
$y = a\left( {1 - {{\left( {\dfrac{x}{a}} \right)}^2}} \right)$
Simplifying it further we get:-
$y = a\left( {1 - \dfrac{{{x^2}}}{{{a^2}}}} \right)$
Now taking LCM and then simplifying it further we get:-

$$y = a\left( {\dfrac{{{a^2} - {x^2}}}{{{a^2}}}} \right) \\\ \Rightarrow y = \left( {\dfrac{{{a^2} - {x^2}}}{a}} \right) \\\$$

Now on cross multiplying we get:-

$$ay = {a^2} - {x^2} \\\ \Rightarrow {x^2} = {a^2} - ay \\\$$

Now taking –a as common from right hand side we get:-
${x^2} = - a\left( {y - a} \right)$
which is the required equation of parabola
Hence proved.

Note: The students might make the mistake of forgetting any of the coordinates of the endpoints of the latus rectum of the ellipse which may lead to failure of the proof.
Also, students should note that the following formula is applicable only if $a > b$:
${b^2} = {a^2}\left( {1 - {e^2}} \right)$