Question

The point of intersection of the tangents at the point P on the ellipse$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and its corresponding point Q on the auxiliary circle, lies on the line

• A. $x=\frac{a}{e}$
• B. $x=0$
• C. $y=0$
• D. $Noneofthese$

Answer 1

The correct option is C

$y=0$

Let the point P be $(acos\theta ,bsin\theta )$ with eccentric angle $\theta$. The corresponding point Q on the auxiliary circle will be $(acos\theta ,bsin\theta )$.We know that

the equation of auxiliary circle of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $x^2+y^2=a^2$ we will now find the tangents and find their intersection.

Tangent to ellipse:

$(acos\theta ,bsin\theta )$ is the point ...(1)

tangent to circle:

$(acos\theta ,asin\theta )$ is the point

$x^2+y^2=a^2$ or $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is the circle. ...(2)

(1)-(2) $\Rightarrow$

$\Rightarrow y=0$

So all the tangents intersect on x - axis or y = 0.

$\Rightarrow$ Option (C).