Question

Chords of the ellipse touch the parabola $a{{y}^{2}}=-2{{b}^{2}}x$; prove that locus of their poles is parabola $a{{y}^{2}}=2{{b}^{2}}x$.

Answer 1

In this question, we have to prove the standard equation of polar with respect to the given pole. Thus, we will apply the ellipse-parabola formula to get the solution. Now, we will first use the ellipse equation and then put the coordinates (h,k) in the equation, then we will make the necessary calculations. After that, we will apply the tangent formula, which passes through the parabola. In the last, we will compare both the equations and make the mathematical calculations, to get the required solution for the problem.

Complete step by step answer:
According to the question, we have to prove that the locus of the pole of the parabola is equal to $a{{y}^{2}}=2{{b}^{2}}x$ .
Thus, we will use the ellipse-parabola formula to get the solution.
According to the problem it is given that the ellipse touch the parabola $a{{y}^{2}}=-2{{b}^{2}}x$ , thus its figure is as follows,

Let us first consider the equation of the ellipse, that is
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1(a>b)$
Let us suppose the coordinates of any pole of the chord in ellipse is (h, k). Thus, the above equation changes to,
$\dfrac{hx}{{{a}^{2}}}+\dfrac{ky}{{{b}^{2}}}=1(a>b)$
So, on taking the least common multiple on the left-hand side in the above equation, we get
$\Rightarrow \dfrac{{{b}^{2}}hx+{{a}^{2}}ky}{{{a}^{2}}{{b}^{2}}}=1$
So, we will multiply ${{a}^{2}}{{b}^{2}}$ on both sides in the above equation, we get
$\Rightarrow \dfrac{{{b}^{2}}hx+{{a}^{2}}ky}{{{a}^{2}}{{b}^{2}}}.{{a}^{2}}{{b}^{2}}=1.{{a}^{2}}{{b}^{2}}$
On further solving, we get
$\Rightarrow {{b}^{2}}hx+{{a}^{2}}ky={{a}^{2}}{{b}^{2}}$
So, we will subtract ${{a}^{2}}{{b}^{2}}$ on both sides in the above equation, we get
$\Rightarrow {{b}^{2}}hx+{{a}^{2}}ky-{{a}^{2}}{{b}^{2}}={{a}^{2}}{{b}^{2}}-{{a}^{2}}{{b}^{2}}$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow {{b}^{2}}hx+{{a}^{2}}ky-{{a}^{2}}{{b}^{2}}=0$ -------- (1)
Now, we know that the equation of parabola is $a{{y}^{2}}=-2{{b}^{2}}x$ , thus when the tangent passes through the parabola, the equation of tangent is equal to ,
$y=mx-\dfrac{{{b}^{2}}}{2am}$
Now, we will subtract y on both sides in the above equation, we get
$\Rightarrow y-y=mx-\dfrac{{{b}^{2}}}{2am}-y$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow mx-\dfrac{{{b}^{2}}}{2am}-y=0$
Therefore, we get
$\Rightarrow mx-y-\dfrac{{{b}^{2}}}{2am}=0$ --------- (2)
On comparing equation (1) and (2), we get
$\Rightarrow \dfrac{{{b}^{2}}h}{m}=+\dfrac{{{a}^{2}}k}{-1}=-\dfrac{{{a}^{2}}{{b}^{2}}}{\dfrac{-{{b}^{2}}}{2am}}=0$
On further solving the above equation, we get
$\Rightarrow \dfrac{{{b}^{2}}h}{m}=-\dfrac{{{a}^{2}}k}{1}=+\dfrac{2{{a}^{3}}m}{1}=0$ --------- (3)
Now, on solving the first term and the second term of the above equation, we get
$\Rightarrow \dfrac{{{b}^{2}}h}{m}=-\dfrac{{{a}^{2}}k}{1}$
Thus, we will multiply m on both sides in the above equation, we get
$\Rightarrow \dfrac{{{b}^{2}}h}{m}.m=-\dfrac{{{a}^{2}}k}{1}.m$
On further simplification, we get
$\Rightarrow {{b}^{2}}h=-\dfrac{{{a}^{2}}k}{1}.m$
Now, we will divide $-{{a}^{2}}k$ on both sides in the above equation, we get
$\Rightarrow \dfrac{{{b}^{2}}h}{-{{a}^{2}}k}=-\dfrac{{{a}^{2}}k}{-{{a}^{2}}k}.m$
Therefore, we get
$\Rightarrow m=-\dfrac{{{b}^{2}}h}{{{a}^{2}}k}$ --------- (4)
Now, we will solve the second term and the last term of equation (3), we get
$\Rightarrow -{{a}^{2}}k=2{{a}^{3}}m$
Thus, we will divide $-{{a}^{2}}$ on both sides in the above equation, we get
$\Rightarrow \dfrac{-{{a}^{2}}k}{-{{a}^{2}}}=\dfrac{2{{a}^{3}}m}{-{{a}^{2}}}$
On further simplification, we get
$\Rightarrow k=-2am$
Now, we will put the value of equation (4) in the above equation, we get
$\Rightarrow k=-2a\left( -\dfrac{{{b}^{2}}h}{{{a}^{2}}k} \right)$
On further solving, we get
$\Rightarrow k=-2\left( -\dfrac{{{b}^{2}}h}{ak} \right)$
Now, on multiplying ak on both sides in the above equation, we get
$\Rightarrow k.ak=-2\left( -\dfrac{{{b}^{2}}h}{ak} \right).ak$
Therefore, we get
$\Rightarrow a{{k}^{2}}=2{{b}^{2}}h$
In the last, we will put k=y and h=x in the above equation, we get
$\Rightarrow a{{y}^{2}}=2{{b}^{2}}x$
Therefore, we have proved that the locus of the poles is a parabola $a{{y}^{2}}=2{{b}^{2}}x$.

Note: While solving this problem, always remember the direct formula of writing the chord or polar equation with the help of pole. Always remember the equation of ellipse and the tangent equation that passes through the parabola. Make the accurate mathematical calculations to solve the equation.