JEE MAIN - Mathematics (2005 - No. 50)
Let $$P$$ be the point $$(1, 0)$$ and $$Q$$ a point on the parabola $${y^2} = 8x$$. The locus of mid point of $$PQ$$ is :
$${y^2} - 4x + 2 = 0$$
$${y^2} + 4x + 2 = 0$$
$${x^2} + 4y + 2 = 0$$
$${x^2} - 4y + 2 = 0$$
Explanation
$$P = \left( {1,0} \right)\,\,Q = \left( {h,k} \right)$$ Such that $${k^2} = 8h$$
Let $$\left( {\alpha ,\beta } \right)$$ be the midpoint of $$PQ$$
$$\alpha = {{h + 1} \over 2},\,\,\,\beta = {{k + 0} \over 2}$$
$$ \therefore $$ $$2\alpha - 1 = h\,\,\,\,\,\,2\beta = k.$$
$${\left( {2\beta } \right)^2} = 8\left( {2\alpha - 1} \right) \Rightarrow {\beta ^2} = 4\alpha - 2$$
$$ \Rightarrow {y^2} - 4x + 2 = 0.$$
Let $$\left( {\alpha ,\beta } \right)$$ be the midpoint of $$PQ$$
$$\alpha = {{h + 1} \over 2},\,\,\,\beta = {{k + 0} \over 2}$$
$$ \therefore $$ $$2\alpha - 1 = h\,\,\,\,\,\,2\beta = k.$$
$${\left( {2\beta } \right)^2} = 8\left( {2\alpha - 1} \right) \Rightarrow {\beta ^2} = 4\alpha - 2$$
$$ \Rightarrow {y^2} - 4x + 2 = 0.$$
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