Question

From the point $(\lambda ,3)$ tangents are drawn to $\frac{x^2}{9}+\frac{y^2}{4}=1$ and are perpendicular to each other then $\lambda =$

• A. $\pm 1$
• B. $\pm 2$
• C. $\pm 3$
• D. $\pm 4$

Answer 1

Answer: (B)

$\pm 2$

We have given equation of ellipse as $\frac{x^2}{9}+\frac{y^2}{4}=1$ and a pair of tangents are drawn from point $(\lambda ,3)$ which are perpendicular to each other.

Now as we know a line $y=mx+c$ will be tangent on ellipse when $c=\pm \sqrt{a^2{m}^2+b^2}$

For this equation $a=3$ and $b=2$ and line is passing through $(\lambda ,3)$ So we can write equation of tangent as :-

$⟹$ $3=m\lambda \pm \sqrt{9m^2+4}$

$⟹$ $3-m\lambda =\pm \sqrt{9m^2+4}$

Now squaring both side of equation we get :-

$⟹$ $9+m^2{\lambda }^2-6m\lambda =9m^2+4$

$⟹$$(9-{\lambda }^2)m^2+6\lambda m-5=0$

So this equation has two roots $m_1,m_2$.

where $m_1{m}_2=$ $\frac{-5}{(9-{\lambda }^2)}$

And as it is given both lines are perpendicular so $m_1\ast m_2=-1$

$⟹$ $\frac{-5}{(9-{\lambda }^2)}$ = $-1$

$⟹(9-{\lambda }^2)=5$

$⟹{\lambda }^2=4$

$⟹\lambda =\pm 2$