The locus of a point, from where tangents to the rectangular hyperbola contain an angle of $45^{0}$ , is

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Solution

The correct option is C
Equation of tangent to the hyperbola
$y = m x \pm \sqrt{m^{2} a^{2} - a^{2}} \Rightarrow L e t P (x_{1}, y_{1}) \Rightarrow y - m x = \pm \sqrt{m^{2} a^{2} - a^{2}} \Rightarrow m^{2} (x_{1}^{2} - a^{2}) - 2 y_{1} x_{1} m = y_{1}^{2} + a^{2} = 0 m_{1} + m_{2} = \frac{2 x_{1} y_{1}}{x_{1}^{2} - a^{2}}; m_{1} m_{2} = \frac{y_{1}^{2} + a^{2}}{x_{1}^{2} - a^{2}} t a n 45^{\circ} = ∣ ∣ \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} ∣ ∣ \Rightarrow (1 + m_{1} m_{2}) = (m_{1} - m_{2}) = (m_{1} + m_{2})^{2} - 4 m_{1} m_{2} \Rightarrow (1 + \frac{y_{1} + a^{2}}{x_{1}^{2} - a^{2}}) = (\frac{2 x_{1} y_{1}}{x_{1}^{2} - a^{2}}) - 4 (\frac{y_{1}^{2} + a^{2}}{x_{1}^{2} - a^{2}})$

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Similar questions

x^{2} - y^{2} = a^{2}

45^{o}

x^{2} - y^{2} = a^{2}

45^{\circ}

x^{2} - y^{2} = a^{2}

45^{\circ}

y^{2} = 4 a x

45^{0}

y^{2} - 4 a x = (x + a)^{2}

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

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