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Question: AB is a focal chord of x<sup>2</sup> – 2x + y – 2 = 0 whose focus is 'S'. If AS = l<sub>1</sub> then...

AB is a focal chord of x2 – 2x + y – 2 = 0 whose focus is 'S'. If AS = l1 then BS is equal to –

A

4l14l1−1\frac{4\mathcal{l}_{1}}{4\mathcal{l}_{1} - 1}

B

l14l1−1\frac{\mathcal{l}_{1}}{4\mathcal{l}_{1} - 1}

C

2l14l1−1\frac{2\mathcal{l}_{1}}{4\mathcal{l}_{1} - 1}

D

None

Answer

l14l1−1\frac{\mathcal{l}_{1}}{4\mathcal{l}_{1} - 1}

Explanation

Solution

x2 – 2x + 1 = – y + 2 + 1

(x – 1)2 = – (y – 3)

Semi latus rectum is equal to the H.M.

2(14)=2×l1×l2l1+l22\left( \frac{1}{4} \right) = \frac{2 \times \mathcal{l}_{1} \times \mathcal{l}_{2}}{\mathcal{l}_{1} + \mathcal{l}_{2}}

l1 + l2 = 4l1l2

l1 = 4l1l2 – l2

BS = l2 = l14l1−1\frac{\mathcal{l}_{1}}{4\mathcal{l}_{1} - 1}