JEE MAIN - Mathematics (2015 (Offline) - No. 14)
Let $$O$$ be the vertex and $$Q$$ be any point on the parabola, $${{x^2} = 8y}$$. If the point $$P$$ divides the line segment $$OQ$$ internally in the ratio $$1:3$$, then locus of $$P$$ is :
$${y^2} = 2x$$
$${{x^2} = 2y}$$
$${{x^2} = y}$$
$${y^2} = x$$
Explanation
Let the coordinates of Q and P be (x1, y1) and (h, k) respectively.
$$\because$$ Q lies on x2 = 8y,
$$\therefore$$ x$$_1^2$$ = 8y ....... (1)
Again, P divides OQ internally in the ratio 1 : 3.
$$\therefore$$ $$h = {{{x_1} + 0} \over 4} = {{{x_1}} \over 4}$$ or x1 = 4h and
$$k = {{{y_1} + 0} \over 4} = {{{y_1}} \over 4}$$ or y1 = 4k
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Now putting x1 and y1 in (1) we get,
16h2 = 32k or, h2 = 2k
$$\therefore$$ the locus of P is given by, x2 = 2y.
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