Question

The normal chord of a parabola $y^2=4ax$ at a point whose ordinate is equal to abscissa, subtends a right angle at

• A. focus
• B. vertex
• C. ends of the latusrectum
• D. any point on directrix

Answer 1

The correct option is A focus

Let $P(a{t}^2,2at)$ be a point on the parabola $y^2=4ax$.
Then, $a{t}^2=2at\Rightarrow t=2$
So the coordinates of $P$ are $(4a,4a)$.
Let $PQ$ be a normal chord normal at $P$ to the parabola.
$y+2x=12a$
Let the coordinates of $Q$ be $(a{t}_1^2,2a{t}_1)$.
$2a{t}_1+2a{t}_1^2=12a\Rightarrow t_1^2+t_1-6=0\Rightarrow t_1=-3[\because t_1\ne 2]$
So, the coordinates of $Q$ are $(9a,-6a)$.
Now, slope of $SP\times$ slope of $SQ$
$=\frac{4a}{3a}\times \frac{-6a}{8a}=-1$
Thus, the normal chord makes a right angle at the focus.