Question

If $P(-3,2)$ is one end of the focal chord $PQ$ of the parabola $y^2+4x+4y=0$, then the slope of the normal at $Q$ is

• A. $\frac{-1}{2}$
• B. $-2$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

The correct option is A $\frac{-1}{2}$

The equation of the parabola is $y^2+4y+4=-4(x-1)\Rightarrow (y+2{)}^2=-4(x-1)$

Compairing with general equation $(y-k{)}^2=4a(x-h)$

ie, vertex$=(h,k)=(1,-2)$,

focal point$=(h+a,k)=(1-1,-2)=(0,-2)$

$P(a{t}_1^2+h+,2a{t}_1+k)=P(-3,2)\Rightarrow t_1=-2$

ie, $t_2=\left(\frac{1}{2}\right)$

$\Rightarrow Q(a{t}_2^2+h,2a{t}_2+k)=Q(-(\frac{1}{2}{)}^2+1,2\frac{-1}{2}-2)=Q(\frac{3}{4},-3)$

On differentiating the equation,

$2y\frac{dy}{dx}+4\frac{dy}{dx}=-4$

Slope of the normal of the equation$=-\frac{dx}{dy}=\frac{2y+4}{4}$

Then Slope of the normal at $Q(\frac{3}{4},-3)=\frac{2y+4}{4}{]}_{(\frac{3}{4},-3)}=\frac{-2}{4}=\frac{-1}{2}$

So, Option A is correct.