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

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

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

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

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

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

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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