Question ID: #602
Let the focal chord $PQ$ of the parabola $y^{2}=4x$ make an angle of $60^{\circ}$ 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 $(0, \alpha)$, then $5\alpha^{2}$ is equal to:
- (1) 15
- (2) 25
- (3) 30
- (4) 20
Solution:
*Property:* The circle described on the focal radius (segment joining focus to a point on the parabola) as diameter touches the tangent at the vertex.
For parabola $y^2 = 4x$, the focus is $S(1, 0)$.
Let point $P$ be $(t^2, 2t)$.
Slope of focal chord $PS$ is $\tan 60^\circ = \sqrt{3}$.
$$ \frac{2t – 0}{t^2 – 1} = \sqrt{3} \Rightarrow 2t = \sqrt{3}(t^2 – 1) $$
$$ \sqrt{3}t^2 – 2t – \sqrt{3} = 0 $$
$$ (t – \sqrt{3})(\sqrt{3}t + 1) = 0 $$
Since $P$ is in the first quadrant ($t > 0$), we get $t = \sqrt{3}$.
Coordinates of $P$:
$$ P(t^2, 2t) = (3, 2\sqrt{3}) $$
Equation of the circle with diameter $PS$:
$$ (x-1)(x-3) + (y-0)(y-2\sqrt{3}) = 0 $$
Since the circle touches the y-axis at $(0, \alpha)$, put $x=0$:
$$ (-1)(-3) + y(y-2\sqrt{3}) = 0 $$
$$ 3 + y^2 – 2\sqrt{3}y = 0 $$
$$ (y – \sqrt{3})^2 = 0 $$
$$ y = \sqrt{3} $$
So, $\alpha = \sqrt{3}$.
$$ 5\alpha^2 = 5(\sqrt{3})^2 = 15 $$
Ans. (1)
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