Question

Q is a point on the auxiliary circle corresponding to the point P of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$. If T is the foot of the

perpendicular dropped from the focus S onto the tangent to the auxiliary circle at Q, then the ∆SPT is

• A. Isosceles
• B. Equilateral
• C. Right angled
• D. Right isosceles

Answer 1

Answer: (A)

Isosceles

Answer 2

Tangent at Q is, x cosθ + y sinθ= a

ST = |ae cos θ - a| = a(1 - e cosθ)

Also, SP = e PM = e $\left( \frac{a}{e} - a\cos\theta \right)$ = a (1 - e cos θ)

⇒ ST = SP, hence isosceles