Question

The tangent from the point $(\lambda ,3)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ are at right angles then $\lambda$ is

• A. $\pm 1$
• B. $\pm 3$
• C. $\pm 2$
• D. none of these

Answer 1

Answer: (C)

$\pm 2$

Let $y=mx+c$ be tangent to the given ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$
The condition for a line $y=mx+c$ to become tangent to the standard ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
$c^2=a^2{m}^2+b^2$
$\Rightarrow c^2=9m^2+4$ and as line passes through $(\lambda ,3)$ i.e $3=m\lambda +c$
eliminating $c$ from the above equations gives
$({\lambda }^2-9)m^2-6\lambda m+5=0$
as tangents are at right angles
$m_1{m}_2=-1\Rightarrow \frac{5}{{\lambda }^2-9}=-1$
$\therefore \lambda =\pm 2$
Alternately, the tangents are perpendicular. Hence, the point must lie on the director circle.
Hence, ${\lambda }^2+3^2=a^2+b^2$
$\Rightarrow {\lambda }^2+9=13$
$\Rightarrow {\lambda }^2=4$
$\Rightarrow \lambda =\pm 2$