PUNE MCA Exam 2016 Question Paper with answers for online practice

Concept:The intersection point of tangents to the parabola $y^2 = 4ax$ satisfies $x_3^2 = x_1 x_2$ and $y_3 = \frac{y_1 + y_2}{2}$.Explanation:Let the points of contact be $(x_1, y_1)$ and $(x_2, y_2)$ on $y^2 = 4ax$, so $y_1^2 = 4ax_1$ and $y_2^2 = 4ax_2$.The equations of tangents at these points are $y y_1 = 2a(x + x_1)$ and $y y_2 = 2a(x + x_2)$.Their intersection $(x_3, y_3)$ satisfies both equations.Subtracting the two tangent equations gives $y_3 (y_1 - y_2) = 2a(x_1 - x_2)$, so $y_3 = \frac{2a(x_1 - x_2)}{y_1 - y_2}$.Substituting $x_1 = \frac{y_1^2}{4a}$ and $x_2 = \frac{y_2^2}{4a}$ yields $y_3 = \frac{y_1 + y_2}{2}$.From the first tangent equation, $x_3 = \frac{y_3 y_1}{2a} - x_1 = \frac{y_1 y_2}{4a}$.Thus $x_3^2 = \left(\frac{y_1 y_2}{4a}\right)^2 = \frac{y_1^2 y_2^2}{16a^2} = \left(\frac{y_1^2}{4a}\right)\left(\frac{y_2^2}{4a}\right) = x_1 x_2$.Therefore the correct relation is $x_3^2 = x_1 x_2$, which corresponds to option A given the likely typographical formatting.Answer: Option A