Question

Tangents are drawn from a point $P\left(6,\sqrt{5}\right)$ to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ touching the ellipse in the points $Q$ and $R$. The angle between $PQ$ and $PR$ is

• A. ${45}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

Answer: (C)

${90}^{\circ }$

Solution -

Let $\theta$ is the angle between the tangents

Let a line $y=mx+c$ is tangent to

the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$

then $c=\sqrt{25m^2+16}$

$y=mx+\sqrt{25m^2+16}$

It passes through $(6,\sqrt{5})$

$(\sqrt{5}-6m{)}^2=(25m^2+16)$

$36m^2+5-12\sqrt{5}m=25m^2+16$

$11m^2-12\sqrt{5}m-11=0$

$m_1{m}_2=\frac{-11}{11}=-1$

Tangents are perpendicular

$\theta ={90}^{\circ }$

$C$ is correct