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Question
Two tangents to the parabola y2 = 8x meet the tangents at the vertex in the point P and Q. If PQ = 4, prove that the equation of the locus of the point of intersection of two tangent is y2 = 8(x + 2).
Solution
Given parabola is y2 = 8x
Comparing with y2 = 4ax, we get,
4a = 8
∴ a = 2
Let M(t1) and N(t2) be any two points on the parabola.
The equations of tangents at M and N are
`"yt"_1 = "x" + 2"t"_1^2` ...(1)
`"yt"_2 = "x" + 2"t"_2^2` ...(2) ...[∵ a = 2]
Let tangent at M meet the tangent at the vertex in P.
But tangent at the vertex is Y-axis whose equation is x = 0.
∴ to find P, put x = 0 in (1)
∴ yt1 = `2"t"_1^2`
∴ y = 2t1 ...(t1 ≠ 0 otherwise tangent at M will be x = 0)
∴ P = (0, 2t1)
Similarly, Q = (0, 2t2)
It is given that PQ = 4
∴ |2t1 – 2t2| = 4
∴ |t1 – t2| = 2 ...(3)
Let R = (x1, y1) be any point on the required locus
Then R is the point of intersection of tangents at M and N.
To find R, we solve (1) and (2).
Subtracting (2) from (1), we get
y(t1 – t2) = `2"t"_1^2-2"t"_2^2` = 2(t1 – t2)(t1 + t2)
∴ y = 2(t1 + t2) ...[∵ M, N are distinct ∴ t1 ≠ t2]
i.e., y1 = 2(t1 + t2) ...(4)
∴ from (1), we get
2t1(t1 + t2) = x + `2"t"_1^2`
∴ 2t1t2 = x i.e. x1 = 2t1t2 ...(5)
To find the equation of locus of R(x1, y1), we eliminate t1 and t2 from the equations (3), (4) and (5).
We know that,
(t1 + t2)2 = (t1 − t2)2 + 4t1t2
∴ `((y_1)/2)^2 = 4 + 4((x_1)/2)` ...[By (3), (4) and (5)]
∴ `"y"_1^2` = 16 + 8x1 = 8(x1 + 2)
Replacing x1 by x and y1 by y, the equation of required locus is y2 = 8(x + 2).
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