Question

Let $A(9,6)$ be the fixed point on the parabola $y^2=4x$. If the perpendicular chords are drawn through this point $A$, then the chord joining the other extremities passes through

• A. $(13,3)$
• B. $(1,3)$
• C. $(13,6)$
• D. $(13,-6)$

Answer 1

Answer: (D)

$(13,-6)$

Let $P(t_1^2,2t_1)$ and $Q(t_2^2,2t_2)$ be the other extremes of the perpendicular chords through $A(9,6)$

The equation of the chord $PQ$ is
$y(t_1+t_2)=2x+2t_1{t}_2$ $....\left(1\right)$

For $A,t=3$ $(\because 9=3^2$ and $6=2\times 3)$
$\therefore$ Slope of $AP=\frac{2}{t_1+3}$

Slope of $AQ=\frac{2}{t_2+3}$

Since $AP$ and $AQ$ are perpendicular
$\Rightarrow \frac{4}{(t_1+3)(t_2+3)}=-1$
$\Rightarrow t_1{t}_2+3(t_1+t_2)+9=-4$
$\Rightarrow 26+2t_1{t}_2=-6(t_1+t_2)$ $....\left(2\right)$

Comparing $\left(2\right)$ with $\left(1\right),$ we get
$\Rightarrow (13,-6)$ for all values of $t_1,t_2$

Hence, option D.