Question

The tangent at any point on the ellipse 16x2 + 25y2 = 400 meets the tangents at the ends of the major axis at T1 and T2. The circle on T1T2 as diameter passes through-

• A. (3, 0)
• B. (0, 0)
• C. (0, 3)
• D. (4, 0)

Answer 1

Answer: (A)

(3, 0)

Answer 2

$\frac{x^{2}}{5^{2}} + \frac{y^{2}}{4^{2}}$= 1

Any tangent to the ellipse is $\frac{x\cos\theta}{5} + \frac{y\sin\theta}{4}$ = 1

This meets x = a = 5 at T1 $\left\{ 5,\frac{4}{\sin\theta}(1–\cos\theta) \right\}$

=$\left\{ 5,4\tan{}\frac{\theta}{2} \right\}$ and meets x = – a = – 5 at

T2 $\left\{ –5,\frac{4}{\sin\theta}(1 + \cos\theta) \right\}$ = $\left\{ –5,4\cot\frac{\theta}{2} \right\}$

The circle on T1, T2 as diameter is (x – 5) (x + 5) +

$\left( y–4\cot\frac{\theta}{2} \right)$ = 0

x2 + y2 – 4y $\left( \tan\frac{\theta}{2} + \cot\frac{\theta}{2} \right)$ – 25 + 16 = 0

This is obviously satisfied by (3, 0)