## Question

### Solution

Correct option is

t3t’ = –1

Any point P on xy = c2         ….. (1) Similarly another point Q is . Diff. (1) w.r.t. x, So equation of normal at P is ⇒ t3x – ty + (c – ct4) = 0                          …. (2)

Equation of line PQ is y – y1 Now (2) and (3) represent same line ⇒ x + ytt’ – c(t + t’) = 0                       …. (3) ⇒ t3t’ = –1

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