Let the focal chord PQ of the parabola 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, ), then is equal to

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Solution

Q.
Let the focal chord PQ of the parabola $y^{2} = 4 x$ 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 $5 α^{2}$ is equal to : [2025]

1 15

2 25

3 20

4 30

Ans.
(1)

Let the coordinate of a point P be $(t^{2}, 2 t)$

Coordinate of focus S of the parabola is (1, 0).

Now, $\tan 60 ° = \frac{2 t - 0}{t^{2} - 1} = \sqrt{3} \Rightarrow \sqrt{3} t^{2} - 2 t - \sqrt{3} = 0$

$\Rightarrow (\sqrt{3} t + 1) (t - \sqrt{3}) = 0 \Rightarrow t = \sqrt{3}$ { $∵$ P lies in first quadreant}

$∴ P (3, 2 \sqrt{3})$

Equation of circle is $(x - 1) (x - 3) + (y - 0) (y - 2 \sqrt{3}) = 0$

At x = 0, $3 + y^{2} - 2 \sqrt{3} y = 0$

$\Rightarrow {(y - \sqrt{3})}^{2} = 0 \Rightarrow y = \sqrt{3}$

$\Rightarrow y = \sqrt{3} = α$

$∴ 5 α^{2} = 15$ .

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