Question

A tangent to the ellipse $x^2+4y^2=4$ meets the ellipse $x^2+2y^2=6$ at P and Q the tangents at P and Q of the ellipse $x^2+2y^2=6$ is

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

Answer 1

The correct option is A ${90}^o$

$x^2+4y^2=4$

$\frac{x^2}{4}+\frac{y^2}{1}=1$

$x^2+2y^2=6$

$\frac{x^2}{6}+\frac{y^2}{3}=1$

Equation of tangent = $\frac{xcos\theta }{2}+\frac{ysin\theta }{1}=1$

chord $=\frac{hx}{6}+\frac{ky}{3}=1$

$\frac{cos\theta }{h/3}=\frac{sin\theta }{k/3}=1\Rightarrow co{s}^2\theta =\frac{h^2}{9}si{n}^2\theta =\frac{k^2}{9}$

$1=\frac{h^2}{9}+\frac{k^2}{9}\Rightarrow h^2+k^2=9=6+3=a^2+b^2$

angle between tangent ${90}^{\circ }$