Question

A tangent to the ellipse $\frac{x^{2}}{9}$+ $\frac{y^{2}}{4}$ = 1 is cut by the tangent at the extremities of the major axis at T and T' and the circle on TT' as diameter passes through the point Q, then Q may be –

• A. (–$\sqrt{5}$, 0)
• B. (2, 3)
• C. (0, 0)
• D. (3, 2)

Answer 1

Answer: (A)

(–$\sqrt{5}$, 0)

Answer 2

$\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$

equation of tangent

$\frac{x(3\cos\theta)}{9}$ + $\frac{y(2\sin\theta)}{4}$ = 1

$\frac{x\cos\theta}{3}$ + $\frac{y\sin\theta}{2}$ = 1

T $\left( 3,\frac{2(1–\cos\theta)}{\sin\theta} \right)$, $T'\left( –3,\frac{2(1 + \cos\theta)}{\sin\theta} \right)$

equation of circle TT' as diameter

(x+3) (x – 3) + $\left( y–\frac{2(1–\cos\theta)}{\sin\theta} \right)$ $\left( y–\frac{2(1 + \cos\theta)}{\sin\theta} \right)$ = 0

x² – 9 + y² + $\frac{4}{\sin\theta}$sin²q – 4cosecq y = 0

x² + y² – 4y cosecq – 5 = 0

which satisfied by 1^st option only.