## Question

### Solution

Correct option is

xy = 2c2

x – a = 0                                                        ……. (1)

y – b =  0                                                       ……. (2)

are conjugate w.r.t.

Let pole of (1) be (x1y1) w.r.t.                       ....... (3)

Then its polar is x1y + xy= 2c2                      …… (4)

Compare (1) and (4) Now line (1) and (2) are conjugate and so pole of one line lies on the other line

⇒  (x1y1) lies on (2)         ⇒ y1 – b =0 ∴ locus of (a, b) is xy = 2c2

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