%%

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 :

JEE · Math · previous-year question

  1. A.$${y^2} - 4x + 2 = 0$$correct
  2. B.$${y^2} + 4x + 2 = 0$$
  3. C.$${x^2} + 4y + 2 = 0$$
  4. D.$${x^2} - 4y + 2 = 0$$

Answer

A. $${y^2} - 4x + 2 = 0$$

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 $$PQ$$ $$\alpha = {{h + 1} \over 2},\,\,\,\beta = {{k + 0} \over 2}$$ $$ \therefore $$ $$2\alpha - 1 = h\,\,\,\,\,\,2\beta = k.$$ $${\left( {2\beta } \right)^2} = 8\left( {2\alpha - 1} \right) \Rightarrow {\beta ^2} = 4\alpha - 2$$ $$ \Rightarrow {y^2} - 4x + 2 = 0.$$

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions