Question

The locus of point of intersection of tangents at the end of normal chord of hyperbola $x^2-y^2=a^2$ is

• A. $a^2(y^2-x^2)=4x^2{y}^2$
• B. $a^2(y^2+x^2)=4x^2{y}^2$
• C. $y^2+x^2=4a^2{x}^2$
• D. $y^2-x^2=4a^2{x}^2$

Answer 1

The correct option is A $a^2(y^2-x^2)=4x^2{y}^2$

Let the point of intersection be P(x,y), then the equation of hyperbola becomes

$x{x}_1-y{y}_1=a^2$

which subtend angle $\theta$ such that we have $(asec\theta ,atan\theta ),$ then

$\frac{x}{sec\theta }+\frac{y}{tan\theta }=2a$

then $\frac{x_1}{\frac{1}{sec\theta }}+\frac{y_1}{\frac{1}{tan\theta }}=\frac{a}{2}$

$sec\theta =\frac{a}{2x_1},tan\theta =\frac{-a}{2y_1}$

Since, ${sec}^2\theta -{tan}^2\theta =1$

putting the values here, we have

$⟹\frac{a^2}{2{x_1}^2}-\frac{a^2}{2{y_1}^2}=1$

$⟹4a^2{y_1}^2-4{x_1}^2{a}^2=16{x_1}^2{y_1}^2$

$⟹a^2({y_1}^2-{x_1}^2)=4{x_1}^2{y_1}^2$

Therefore we get the equation as

$⟹a^2(y^2-x^2)=4x^2{y}^2$ (Answer)

Therefore option(A )is correct.