Question

Tangents one to each of the ellipses $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{a^2+\lambda }+\frac{y^2}{b^2+\lambda }=1$ are drawn. If the tangents meet at right angles, then the locus of their point of intersection is:

• A. $x^2+y^2=a^2+\lambda$
• B. $x^2+y^2=b^2+\lambda$
• C. $x^2+y^2=a^2+b^2+\lambda$
• D. $x^2+y^2=a^2{x}^2-b^2+\lambda$

Answer 1

The correct option is C $x^2+y^2=a^2+b^2+\lambda$

Let intersection of two tangents be $(x_0,y_0)$

$\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}=1$ ......... $\left(1\right)$

$\frac{x_0^2}{a^2+\lambda }+\frac{y_0^2}{b^2+\lambda }=1$ ......... $\left(2\right)$

Slope at $(x_0,y_0)$ to both the tangents are :

$m_1=-\frac{x_0{b}^2}{y_0{a}^2}$

$m_2=-\frac{x_0(b^2+\lambda )}{y_0(a^2+\lambda )}$

Now, $m_1.m_2=-1$

$\frac{x_0{b}^2}{y_0{a}^2}.\frac{x_0(b^2+\lambda )}{y_0(a^2+\lambda )}=-1$

$x_0^2{b}^2(b^2+\lambda )+y_0^2{a}^2(a^2+\lambda )=0$

From eq-1 we get,

$x_0^2{b}^2+y_0^2{a}^2=a^2{b}^2$

From eq-2 we get,

$x_0^2{b}^2+y_0^2{a}^2+\lambda x_0^2+\lambda y_0^2=(a^2+\lambda )(b^2+\lambda )$

$\Rightarrow x_0^2+y_0^2=a^2+b^2+\lambda$

$\Rightarrow x^2+y^2=a^2+b^2+\lambda$