Question

Prove that the common tangent of the ellipses $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{2x}{c}$ and $\frac{x^2}{b^2}+\frac{y^2}{a^2}+\frac{2x}{c}=0$ subtends a right angle t the origin.

Answer 1

$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2x}{c}=0$ ---eq.1

$\frac{x^2}{b^2}+\frac{y^2}{a^2}-\frac{2x}{c}=0$ ---eq.1

Multiply eq.1 $\times a^2$

$x^2+\frac{y^2.a^2}{b^2}-\frac{2a^{2}x}{c}=0$

${(x-\frac{a^2}{c})}^2-\frac{a^4}{c^2}+\frac{a^2}{b^2}{y}^2=0$

${(x-\frac{a^2}{c})}^2+\frac{a^2}{b^2}{y}^2=\frac{a^4}{c^2}$

$\frac{(x-a^2/c^2{)}^2}{1}+\frac{y^2}{b^2/a^2}=\frac{a^4}{c^2}$

$\frac{(x-a^2/c^2{)}^2}{(a/c{)}^2}+\frac{y^2}{(ba/c{)}^2}=1$ ---eq.3

Multiply eq.2 $\times b^2$

$x^2+\frac{y^2.b^2}{a^2}-\frac{2b^{2}x}{c}=0$

${(x-\frac{b^2}{c})}^2-\frac{b^4}{c^2}+\frac{b^2}{a^2}{y}^2=0$

${(x-\frac{b^2}{c})}^2+\frac{b^2}{a^2}{y}^2=\frac{b^4}{c^2}$

$\frac{(x-b^2/c^2{)}^2}{1}+\frac{y^2}{a^2/b^2}=\frac{b^4}{c^2}$

$\frac{(x-b^2/c^2{)}^2}{(b/c{)}^2}+\frac{y^2}{(ab/c{)}^2}=1$ ---eq.4

We know,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Tangent equation in slope form

$y=mx\pm \sqrt{a^2{m}^2+b^2}$

$ \therefore C_1 \left( \dfrac{a^2}{c} , o \right) \, \, C_2 \left( \dfrac{b^2}{c} , o \right) $

Therefore,

$y=m_1(x-\frac{a^2}{c})\pm \sqrt{\frac{a^4}{c^2}{m}_1^2+\frac{a^2{b}^2}{c^2}}$

$y=m_1(x-\frac{a^2}{c})\pm \frac{a}{c}\sqrt{a^2{m}_1^2+b^2}$ ---eq.5

Again,

$y=m_2(x-\frac{b^2}{c})\pm \sqrt{\frac{b^4}{c^2}{m}_2^2+\frac{a^2{b}^2}{c^2}}$

$y=m_2(x-\frac{b^2}{c})\pm \frac{b}{c}\sqrt{b^2{m}_1^2+a^2}$ ---eq.6

eq.5 and eq.6 passing by origin

$\therefore \frac{m_1{a}^2}{c}=\pm \frac{a}{c}\sqrt{a^2{m}_1^2+b^2}$

$\frac{m_2{b}^2}{c}=\pm \frac{b}{c}\sqrt{b^2{m}_2^2+a^2}$

$m_{1}a=\pm \sqrt{a^2{m}_1^2+b^2}$ ---eq.7

$m_{2}b=\pm \sqrt{b^2{m}_2^2+a^2}$

In eq. 7 , squaring both sides.

$m_1^2{a}^2=a^2{m}_1^2+b^2{b}^2=0b=0$

So, $m_2=\pm \frac{\sqrt{b^2{m}_2^2+a^2}}{b}=\pm \frac{0+a^2}{0}=\frac{a}{0}=\infty$

and $m_1=0$

So, the tangents are perpendicular.