tan60∘=t2−12t−0=3⇒t=3
∴P(3,23)
Circle :
(x−1)(x−3)+(y−0)(y−23)=0
at x=0
⇒3+y2−23y=0
⇒y=3=α
5α2=15
Let the focal chord PQ of the parabola y2=4x make an angle of 60∘ with the positive x-axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, S being the focus of the parabola, touches the y-axis at the point (0,α), then 5α2 is equal to :
Held on 2 Apr 2025 · Verified 6 Jul 2026.
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