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Question

Find the locus of point of intersection of tangents to the parabola y2=4ax which include an angle α

A
(xa)2tan2α=y24ax
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B
(x+a)2sec2α=y24ax
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C
(xa)2sec2α=y24ax
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D
(x+a)2tan2α=y24ax
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Solution

The correct option is D (x+a)2tan2α=y24ax
Let the tangents be drawn at the points t1 and

t2
and (h,k) be their point of intersection.

h=at1t2,k=a(t1+t2) ...(1)

Also, their equations are t1y=x+at21,t2y=x+at22
Also their slope are 1t1 and 1t2 ...(2)
If they include an angle α, then
tanα=(t2t1)t1t2+1
or tan2α(1+t1t2)2={(t1+t2)24t1t2}
or tan2α(1+ha)2={k2a24ha}
or tan2α(h+a)2=(k24ah)
Hence, the required locus is(x+a)2tan2α=y24ax.

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