The correct option is
D π2
Point whose ordinate equal to abscissa will be such that x=y,
but (x,y) lies on y2=4ax
⇒ x2=4ax
⇒ x=0 or x=4a
Since at (0,0) we cannot draw a normal chord, hence point at which normal chord is drawn is P(4a,4a).
Focus of the given parabola y2=4ax is S(a,0).
Equation of the normal chord at (4a,4a) will be
y−4ax−4a=−2
⇒ 2x+y=12a is the equation of normal.
To find the points of intersection of normal with the parabola , we need to
solve the equation (12a−2x)2=4ax
⇒ (6a−x)2=ax
⇒ 36a2−12ax+x2=ax
⇒ x2−13ax+36a2=0
⇒ (x−4a)(x−9a)=0
⇒ x=4a or 9a
P(4a,4a) and Q(9a,−6a) are the end points of the normal chord.
Slope of line PS=(4a−0)(4a−a)=43
Slope of line QS=(−6a−0)(9a−a)=−34
We can observe that product of (SlopeofPS)×(SlopeofQS)=−1
⇒ ∠PSQ=90o.
∴ Normal chord at the point whose ordinate is equal to abscissa subtends an angle of π2 at the focus.