Let the shortest distance from (a, 0), a > 0, to the parabola be 4. Then the equation of the circle passing through the point (a, 0) and the focus of the parabola, and having its centre on the axis of the parabola is: [2025]

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Solution

Q.
Let the shortest distance from (a, 0), a > 0, to the parabola $y^{2} = 4 x$ be 4. Then the equation of the circle passing through the point (a, 0) and the focus of the parabola, and having its centre on the axis of the parabola is: [2025]

1 $x^{2} + y^{2} - 4 x + 3 = 0$

2 $x^{2} + y^{2} - 8 x + 7 = 0$

3 $x^{2} + y^{2} - 10 x + 9 = 0$

4 $x^{2} + y^{2} - 6 x + 5 = 0$

Ans.
(4)

Equation of normal at $P (t^{2}, 2 t)$ is given by

$y + t x = 2 t + t^{3}$ ... (i)

Put x = a, y = 0 in equation (i), we get

$a t = 2 t + t^{3}$

$\Rightarrow a = 2 + t^{2}$

$∴$ The point Q is $(2 + t^{2}, 0)$

$\Rightarrow P Q = 4 \Rightarrow 4 + 4 t^{2} = 16$

$\Rightarrow 4 t^{2} = 12 \Rightarrow t^{2} = 3$

$∴ a = 5 and Q \equiv (5, 0)$

$∵$ Focus of parabola is (1, 0) and centre of circle lie on axis of parabola.
$\Rightarrow$ (1, 0) and (5, 0) will be the end points of diameter of the circle.

$∴$ Equation of circle is $(x - 1) (x - 5) + y^{2} = 0$

$\Rightarrow x^{2} + y^{2} - 6 x + 5 = 0$ .

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