i want the solution of q. 9thanks

Dear student,

The equation of the tangent at P(x1, y1)
(xx1/a²) + (yy1/b²) = (x1²/a²) + (y1²/b²)

And the other tangent at Q(x2, y2)
(xx2/a²) + (yy2/b²) = 1

The point T(x0, y0) belongs to both of these tangents, so:
(x0x1/a²) + (y0y1/b²) = 1

and:
(x0x2/a²) + (y0y2/b²) = 1

The next step is to simply state that the equation of the chord of contact is:
(xx0/a²) + (yy0/b²) = 1

Now use the distance formula from (0,0) ie origin....

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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