Let P be a variable point. From $P, P Q$ and $P R$ are tangents drawn to the parabola $y^{2} = 4 x$ . If $∠ Q P R$ is always $45^{\circ}$ , then the locus of $P$ is

x^{2} - y^{2} + 6 x + 1 = 0

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x^{2} + y^{2} + 6 x + 1 = 0

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x^{2} - y^{2} - 6 x + 1 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} - 6 x + 1 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is A $x^{2} - y^{2} + 6 x + 1 = 0$
$y = m x + \frac{1}{m}$ is equation of tangent to a parabola $y^{2} = 4 x$ with slope $m$ .
$\Rightarrow m^{2} x - m y + 1 = 0$
If $m_{1}$ and $m_{2}$ are roots of this equation then
$m_{1} + m_{2} = \frac{y}{x}; m_{1} m_{2} = \frac{1}{x}$
$m_{1} - m_{2} = \pm \sqrt{(\frac{y}{x})^{2} - 4 (\frac{1}{x})} = \pm \frac{\sqrt{y^{2} - 4 x}}{x}$
$\Rightarrow t a n 45^{0} = 1 = \pm \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}$
$\Rightarrow 1 + \frac{1}{x} = \pm \frac{\sqrt{y^{2} - 4 x}}{x}$
$x^{2} + 1 + 2 x = y^{2} - 4 x$
$x^{2} + 1 + 6 x - y^{2} = 0$
Hence, option 'A' is correct.

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Let P be a variable point. From P,PQ and PR are tangents drawn to the parabola y2=4x. If ∠QPR is always 45∘, then the locus of P is

Let P be a variable point. From $P, P Q$ and $P R$ are tangents drawn to the parabola $y^{2} = 4 x$ . If $∠ Q P R$ is always $45^{\circ}$ , then the locus of $P$ is