Question

If the normal to the parabola $y^2=4x$ at $P(1,2)$ meets the parabola again in $Q$ then $Q=$

• A. $(-6,9)$
• B. $(9,-6)$
• C. $(-9,-6)$
• D. $(-6,-9)$

Answer 1

Answer: (B)

$(9,-6)$

Given, $y^2=4x$ ....... $\left(i\right)$

Differentiating above eqn w.r.t $x$, we get

$2y.y^{\prime }=4$

$\Rightarrow y^{\prime }=\frac{2}{y}$

$\Rightarrow y_{1,2}^{\prime }=1$

Then, the slope of the normal will be $-1$.

And the equation of the normal will be

$\frac{y-2}{x-1}=-1$

$y-2=-x+1$

$x+y=3$

$y=3-x$ ....... $\left(ii\right)$

$y^2=x^2-6x+9$

$4x=x^2-6x+9$ ....... From $\left(i\right)$

$x^2-10x+9=0$

$(x-9)(x-1)=0$

$x=9$ and $x=1$

Therefore, $y=-6,2$ ....... From $\left(ii\right)$

Hence $Q=(9,-6)$.