Question

What is the corresponding point of $(acos\theta ,bsin\theta )$ of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ on its auxiliary cirlcle?

• A. (a cosθ, a sin θ)
• B. (b cosθ, b sin θ)
• C. (a sin θ, b cos θ)
• D. (a sin θ, a cos θ)

Answer 1

The correct option is A

(a cosθ, a sin θ)

Auxiliary circle of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is the circle with its diameter as the major axis of ellipse.

P and Q are known as the corresponding points. Q is the point where perpendicular from P

to x - axis meets the ellipse. If PO makes an angle θ with the major axis (or x - axis in this case) we say that $\theta$ is the eccentric angle of Q. We know that the co-ordinates of P is $(acos\theta ,asin\theta )$.P and Q has the same x-coordinate. If we substitute a $cos\theta$ for x in $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ we will get $y=b$

$sin\theta$. So$(acos\theta ,bsin\theta )$ and$(acos\theta ,asin\theta )$are corresponding points.