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Question

If P(5,1) is one end of the focal chord PQ of the parabola x=y28y+2, then the slope of the tangent at the other end is

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Solution

Tangents at the end of focal chord are perpendicular to each other
Slope of tangent at Q= Slope of normal at P
Slope of normal to y2=4ax at P is t
Given, x=y28y+2
(y4)2=(x+14)
Any point on the parabola will be (14+t24,4+t2)
On comparing it with (5,1), we get t=6
Slope of the normal at P will be =6
Hence, Slope of tangent at Q6

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