Question

Consider the parabola $y^2=4 x$ Let $S$ be the focus of the parabola A pair of tangents drawn to the parabola from the point $P=(-2,1)$ meet the parabola at $P_1$ and $P_2$ Let $Q_1$ and $Q_2$ be points on the lines $S P_1$ and $S P_2$ respectively such that $P Q_1$ is perpendicular to $S P_1$ and $P Q_2$ is perpendicular to $S P_2$ Then, which of the following is/are TRUE ?

• A. $S Q_l=2$
• B. $Q _1 Q _2=\frac{3 \sqrt{10}}{5}$
• C. $PQ _1=3$
• D. $SQ _2=1$

Answer 1

The Correct Option is B, C, D

Let \( P_1 \) and \( P_2 \) be \( (t_1^2, 2t_1) \) and \( (t_2^2, 2t_2) \), respectively, and \( P \equiv (t_1 t_2, t_1 + t_2) \) is given as \( (-2, 1) \). 

\[ t_1 = 2, \, t_2 = -1 \quad \text{or} \quad t_1 = -2, \, t_2 = 2 \] \[ \Rightarrow P_1(4, 4) \, \text{and} \, P_2(1, -2) \]

Slope of \( SP_1 \) is \( \frac{4}{3} \), and the slope of \( SP_2 \) is infinity.

\[ \text{Equation of} \, SP_1: \, y - 0 = \frac{4}{3} (x - 1) \quad \Rightarrow \quad 4x - 3y - 4 = 0 \]

Equation of \( PQ_1 \):

\[ y - 1 = \frac{-4}{3}(x + 2) \quad \Rightarrow \quad 3x + 4y + 2 = 0 \] \[ \Rightarrow Q_1\left( \frac{2}{5}, \frac{-4}{5} \right) \]

Equation of \( SP_2 \) and \( PQ_2 \):

\[ x = 1, \quad y = 1 \quad \Rightarrow Q_2(1, 1) \] \[ \Rightarrow SQ_1, Q_2 = \frac{3\sqrt{10}}{5} \] \[ PQ_1 = \sqrt{ \left( \frac{-2 - 2}{5} \right)^2 + \left( 1 + 4 \right)^2} = \sqrt{\frac{144}{25} + \frac{81}{25}} = \sqrt{\frac{225}{25}} = 3 \] \[ SQ_2 = \sqrt{(1 - 1)^2 + (1 - 0)^2} = \sqrt{1} = 1 \]