Question

The locus of the poles of normal chords of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is

• A. $a^6{y}^2-b^6{x}^2=(a^2+b^2{)}^2{x}^2{y}^2$
• B. $a^4{y}^2-b^4{x}^2=(a^2+b^2{)}^2{x}^2{y}^2$
• C. $a^2{y}^2-b^2{x}^2=(a^2+b^2)x^2{y}^2$
• D. $a^6{y}^2+b^6{x}^2=(a^2+b^2{)}^2{x}^2{y}^2$

Answer 1

The correct option is A $a^6{y}^2-b^6{x}^2=(a^2+b^2{)}^2{x}^2{y}^2$

Equation of any normal to the hyperbola

$\frac{x^2}{a^2}–\frac{y^2}{b^2}=1$.............(1)

$\frac{ax}{\sec \theta }+\frac{by}{\tan \theta }=a^2{+b}^2$............(2)

Let $(x_1,y_1)$ be its pole w.r.t. 1

Polar of $(x_1,y_1)$ w.r.t. 1 is $\frac{x{x}_1}{a^2}–\frac{y{y}_1}{b^2}=1$..........(3)

As (2) and (3) both represent the polar of $(x_1,y_1)$,

comparing coefficient

$\frac{\frac{x_1}{a^2}}{\frac{a}{\sec \theta }}=\frac{–\frac{y_1}{b^2}}{\frac{b}{\tan \theta }}=\frac{1}{a^2+b^2}$

$\sec \theta =\frac{a^3}{x_1{(a}^2+b^2)}$ , $\tan \theta =\frac{b^3}{y_1{(a}^2+b^2)}$

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

$\frac{a^6}{{x_1}^2{(a^2+b^2)}^2}-\frac{b^6}{{y_1}^2{(a^2+b^2)}^2}=1$

Hence locus of $(x_1,y_1)$ is

$\frac{a^6}{x^2}-\frac{b^6}{y^2}={(a^2+b^2)}^2$

$a^6{y}^2-b^6{x}^2={(a^2+b^2)}^2{x}^2{y}^2$

hence proved