From the point $P (α, β)$ , tangents are drawn to the parabola $y^{2} = 4 x$ , including an angle $45^{\circ}$ to each other. Then locus of $P (α, β)$ is

a circle with centre

(- 3, 0)

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an ellipse with centre

(- 3, 0)

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a rectangular hypererbola with centre

(- 3, 0)

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a rectangular hypererbola with centre

(3, 0)

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Solution

The correct option is C a rectangular hypererbola with centre $(- 3, 0)$
Any tangent to $y^{2} = 4 x$ is $y = m x + \frac{1}{m} \Rightarrow α m^{2} - β m + 1 = 0$ (as $P (α, β)$ lies on it)
Here $m_{1} + m_{2} = \frac{β}{α}$ and $m_{1} m_{2} = \frac{1}{α}$
Now, $tan 45^{\circ} = ∣ ∣ ∣ \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} ∣ ∣ ∣$
$∵ (m_{1} - m_{2})^{2} = (m_{1} + m_{2})^{2} - 4 m_{1} m_{2}$
$∴ {(\frac{β}{α})}^{2} - \frac{4}{α} = {(1 + \frac{1}{α})}^{2}$
$\Rightarrow β^{2} - 4 α = (α + 1)^{2} \Rightarrow (α + 3)^{2} - β^{2} = 8$
We have $(x + 3)^{2} - y^{2} = 8$ , which is a rectangular hypererbola with centre $(- 3, 0)$

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