Question

Prove that the chord of contact of tangents drawn from the point (h, k) to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (a > b) will subtend a right angle at the centre if $\frac{h^2}{a^4} + \frac{k^2}{b^4} = \frac{1}{a^2} + \frac{1}{b^2}$.

Answer 1

$\frac{h^2}{a^4} + \frac{k^2}{b^4} = \frac{1}{a^2} + \frac{1}{b^2}$

Answer 2

Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The centre of the ellipse is the origin $O(0, 0)$.
Let the point from which the tangents are drawn be $P(h, k)$.
The equation of the chord of contact of tangents drawn from $P(h, k)$ to the ellipse is given by $T=0$, where $T = \frac{hx}{a^2} + \frac{ky}{b^2} - 1$.
So, the equation of the chord of contact is $\frac{hx}{a^2} + \frac{ky}{b^2} = 1$.

Let the chord of contact intersect the ellipse at points $A$ and $B$. The chord of contact $AB$ subtends a right angle at the centre $O(0, 0)$. This means the lines $OA$ and $OB$ are perpendicular.
The equation of the pair of lines $OA$ and $OB$ can be obtained by homogenizing the equation of the ellipse using the equation of the chord of contact.
The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
From the equation of the chord of contact, we have $1 = \frac{hx}{a^2} + \frac{ky}{b^2}$.
Substitute this expression for 1 into the ellipse equation to make it homogeneous:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = \left(\frac{hx}{a^2} + \frac{ky}{b^2}\right)^2$
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{h^2 x^2}{a^4} + \frac{k^2 y^2}{b^4} + \frac{2hkxy}{a^2 b^2}$
Rearranging the terms to form a homogeneous equation of the second degree in $x$ and $y$:
$\left(\frac{1}{a^2} - \frac{h^2}{a^4}\right) x^2 - \frac{2hk}{a^2 b^2} xy + \left(\frac{1}{b^2} - \frac{k^2}{b^4}\right) y^2 = 0$

This is the equation of the pair of straight lines $OA$ and $OB$. A general homogeneous equation of the second degree $Ax^2 + Bxy + Cy^2 = 0$ represents a pair of straight lines passing through the origin. These lines are perpendicular if and only if the sum of the coefficients of $x^2$ and $y^2$ is zero, i.e., $A+C = 0$.

In our equation, $A = \frac{1}{a^2} - \frac{h^2}{a^4}$ and $C = \frac{1}{b^2} - \frac{k^2}{b^4}$.
Since the lines $OA$ and $OB$ are perpendicular, the condition $A+C=0$ must be satisfied:
$\left(\frac{1}{a^2} - \frac{h^2}{a^4}\right) + \left(\frac{1}{b^2} - \frac{k^2}{b^4}\right) = 0$
$\frac{1}{a^2} - \frac{h^2}{a^4} + \frac{1}{b^2} - \frac{k^2}{b^4} = 0$
Rearranging the terms, we get:
$\frac{h^2}{a^4} + \frac{k^2}{b^4} = \frac{1}{a^2} + \frac{1}{b^2}$

This is the required condition for the chord of contact of tangents drawn from the point $(h, k)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ to subtend a right angle at the centre.
The condition $a > b$ specifies that the major axis is along the x-axis, but it does not affect the derivation of the condition for the chord of contact to subtend a right angle at the centre.