Question

If line $PQ$, whose equation is $y=2x+k,$ is a normal to the parabola whose vertex is $(-2,3)$ and the axis parallel to the $x$-axis with latus rectum equal to $2$, then the possible value of k is

• A. $\frac{58}{8}$
• B. $\frac{50}{8}$
• C. $1$
• D. $-1$

Answer 1

Answer: (C)

$1$

Normal to parabola $\to y=2x+k$

parabola $\to (y-3{)}^2=4a(x+2)$

Latus rectum$=2$

$4a=2$

$a=\frac{1}{2}$

parabola $\to (y-3{)}^2=2(x+2)$

$y=3+\sqrt{2}\sqrt{x+2}$

slope of normal =$\frac{1}{-y^{\prime }}$

$y^{\prime }=\frac{\sqrt{2}}{2\sqrt{x+2}}=\frac{1}{\sqrt{2x+4}}$

$m=-\sqrt{-2x+4}$

if $2=-\sqrt{2x+4}$

then $x=0,y=5,1$

$(0,5$) should also lies on $y=2x+k$

then

$k=5,1$

$C$ is correct