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

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

I want to approach this problem through my method

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ be the equation of the ellipse at $\left( {{x_1},{y_1}} \right)$.

${m_1} = \frac{{{y_1}}}{{{x_1}}} \Rightarrow m{'_1} = - \frac{{{x_1}}}{{{y_1}}}$

${m_2} = \frac{{{y_2}}}{{{x_2}}} \Rightarrow m{'_1} = - \frac{{{x_2}}}{{{y_2}}}$

${x_1}{x_2} + {y_1}{y_2} = 0$ is the condition

$k = - \frac{{{x_1}}}{{{y_1}}}h + \sqrt {\frac{{{a^2}{x_1}^2}}{{{y_1}^2}} + {b^2}} \Rightarrow k = - \frac{{{x_1}}}{{{y_1}}}h + \sqrt {\frac{{{a^2}{y_2}^2}}{{{x_2}^2}} + {b^2}} \Rightarrow k = - \frac{{{x_1}}}{{{y_1}}}h + \sqrt {\frac{{{a^2}{y_2}^2 + {b^2}{x_2}^2}}{{{x_2}^2}}} $

$ \Rightarrow k = - \frac{{{x_1}}}{{{y_1}}}h + \frac{{ab}}{{{x_2}}}\& k = - \frac{{{x_2}}}{{{y_2}}}h + \frac{{ab}}{{{x_1}}}$

On multiplying we get $\Rightarrow {k^2} = \frac{{{x_1}}}{{{y_1}}}\frac{{{x_2}}}{{{y_2}}}{h^2} + \frac{{{a^2}{b^2}}}{{{x_1}{x_2}}} + ab\left( {\frac{1}{{{x_1}}} + \frac{1}{{{x_1}}}} \right)$

$\Rightarrow {k^2} = \frac{{{x_1}}}{{{y_1}}}\frac{{{x_2}}}{{{y_2}}}{h^2} + \frac{{{a^2}{b^2}}}{{{x_1}{x_2}}} - abh\left( {\frac{1}{{{y_1}}} + \frac{1}{{{y_2}}}} \right)$

$ \Rightarrow {k^2} = - {h^2} + \frac{{{a^2}{b^2}}}{{{x_1}{x_2}}} - abh\left( {\frac{1}{{{y_1}}} + \frac{1}{{{y_2}}}} \right)$

How do we proceed form here

Quanto
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Samar Imam Zaidi
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1 Answers1

2

Let $(x,y)$ be the tangential points for the pair of the tangent lines from the point $(h,k)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Then , $$y’=-\frac{b^2x}{a^2y} = \frac{y-k}{x-h}\implies \frac{xh }{a^2}+\frac{yk }{b^2}=1$$ Substitute it into $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and obtain the products below via the Vieta’s formula $$x_1 x_2 = \frac{a^4(b^2-k^2)}{a^2k^2+b^2h^2},\>\>\>\>\>y_1 y_2 = \frac{b^4(a^2-h^2)}{a^2k^2+b^2h^2}$$ Then, the condition ${x_1}{x_2} + {y_1}{y_2} = 0$ for the right angle subtended by the two tangential points leads to

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

Quanto
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