Q.282. Locus of points of intersection of tangents under given conditions.

Question 282.
If normal of the circle x^2 + y^2 +6x +8y +9 =0 intersect the parabola y^2 = 4x at P and Q then find the locus of point of intersection of tangent's at P and Q.

Answer 282.
Let P = (a^2, 2a) and Q = (b^2, 2b) be the points of intersection of the normal of the given circle with the given parabola.

The equations of tangents at P and Q are
2ay = 2(x + a^2) and 2by = 2(x + b^2)
=> ay = x + a^2 and by = x + b^2
Solving the two euqations, their point of intersection is
R = (ab, a+b) = (x, y)

Normal of the circle passes through its centre C(-3, -4)
and P and Q
=> Slope of CP = slope of CQ
=> (2a + 4) / (a^2 + 3) = (2b + 4) / (b^2 + 3)
=> 2ab^2 + 4b^2 + 6a + 12 = 2a^2b + 4a^2 + 6b + 12
=> ab (b - a) + 2(b^2 - a^2) - 3(b - a) = 0
=> ab + 2 (a + b) - 3 = 0.

Plugging variable coordinates of R = (x, y) = (ab, a+b)
=> x + 2y - 3 = 0
which is the required locus.

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