The locus of the point of intersection of tangents to the hyperbola $x^{2} - y^{2} = a^{2}$ which include an angle of 45º is

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

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

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

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

answer is C.

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Detailed Solution

$Let y = mx \pm \sqrt{a^{2} m^{2} - a^{2}} be two tangent s to the hyperbola . If it passes through (x_{1}, y_{1}) then \Rightarrow {(y_{1} - {mx}_{1})}^{2} = a^{2} m^{2} - a^{2} \Rightarrow m^{2} ({x_{1}}^{2} - a^{2}) - 2 {mx}_{1} y_{1} + {y_{1}}^{2} + a^{2} = 0 \Rightarrow m_{1} + m_{2} = \frac{2 x_{1} y_{1}}{{x_{1}}^{2} - a^{2}}, m_{1} m_{2} = \frac{{y_{1}}^{2} + a^{2}}{{x_{1}}^{2} - a^{2}} If' θ' is angle between tangents then \tan 45^{\circ} = \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} \Rightarrow 1 = \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} \Rightarrow m_{1} - m_{2} = 1 + m_{1} m_{2} \Rightarrow {(m_{1} - m_{2})}^{2} = {(1 + m_{1} m_{2})}^{2} \Rightarrow {(m_{1} + m_{2})}^{2} - 4 m_{1} m_{2} = {(1 + m_{1} m_{2})}^{2} \Rightarrow {(\frac{2 x_{1} y_{1}}{{x_{1}}^{2} - a^{2}})}^{2} - 4 (\frac{{y_{1}}^{2} + a^{2}}{{x_{1}}^{2} - a^{2}}) = {(1 + \frac{{y_{1}}^{2} + a^{2}}{{x_{1}}^{2} - a^{2}})}^{2} \Rightarrow 4 {x_{1}}^{2} {y_{1}}^{2} - 4 ({y_{1}}^{2} + a^{2}) ({x_{1}}^{2} - a^{2}) = {({x_{1}}^{2} - a^{2} + {y_{1}}^{2} + a^{2})}^{2} \Rightarrow 4 a^{2} ({y_{1}}^{2} - {x_{1}}^{2} + a^{2}) = {({x^{2}}_{1} + {y_{1}}^{2})}^{2} ∴ l o c u s i s ({x^{2}}_{} + {y_{}}^{2})^{2} = 4 a^{2} (y^{2} - x^{2} + a^{2})$