The correct option is A π2
Equation of a tangent of the parabola y2=4ax having slope m is y=mx+am.
Let there be a point P(−4a, y1) on the line x+4a=0.
If the tangent y=mx+am passes through (−4a, y1), then y1=−4am+am → 4am2+my1−a=0
Let the roots of these equation be m1, m2.
Differentiating the given equation of parabola, we get 2yy'=4a ↔ y=2ay'
y-coordinates of points say A, B from where these tangents are drawn are 2am1, 2am2
We have to find the angle between OA and OB where O is the origin.
y2=4ax → yx=4ay
Slope of a line passing through origin and a point on the parabola(x1, y1) is 4ay1.
Angle between the lines of slopes m1, m2 is tan−1((m1−m21+m1m2∣∣).
Here, slopes are 4a2am1=2m1, 4a2am2=2m2.
So, the angle subtended is tan−1((2m1−2m21+4m1m2∣∣).
m1m2=−14 (Product of the roots)
Denominator becomes zero.
So, the angle subtended is 900.