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Question: Tangent are drawn from the points on the line x – y – 5 = 0 to x<sup>2</sup> + 4y<sup>2</sup> = 4, ...

Tangent are drawn from the points on the line

x – y – 5 = 0 to x2 + 4y2 = 4, then all the chords of contact pass through a fixed point, whose coordinate are-

A

(45,−15)\left( \frac{4}{5}, - \frac{1}{5} \right)

B

(45,15)\left( \frac{4}{5},\frac{1}{5} \right)

C

(−45,15)\left( - \frac{4}{5},\frac{1}{5} \right)

D

None of these

Answer

(45,−15)\left( \frac{4}{5}, - \frac{1}{5} \right)

Explanation

Solution

Let A(x1, x1 – 5) be a point on x – y – 5 = 0, then chord of contact of x2 + 4y2 = 4 w.r.t A is xx1 + 4y (x1 – 5) = 4

Ž (x + 4y) x1 – (20y + 4) = 0

It is passes through a fixed point.

\ x + 4y = 0

and 20y + 4 = 0 (Q from P + lQ = 0)

Ž y = – 15\frac{1}{5}

and x = 45\frac{4}{5}.

The coordinates of fixed point are(45,–15)\left( \frac{4}{5},–\frac{1}{5} \right).

Hence (1) is the correct answer.