Question

Prove that b square x square minus A square y square equals to a square b square if X equals to AC theta y equals to B tan theta

Answer 1

$\textsf{Given:}$

$\mathsf{x=a\;sec\theta}$

$\mathsf{y=b\;tan\theta}$

$\implies\mathsf{\frac{x}{a}=sec\theta\;and\;\frac{y}{b}=tan\theta}$

$\textsf{Using}$

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

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

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

$\implies\mathsf{b^2x^2-a^2y^2=a^2b^2}$

Answer 2

Answer:

⟹ax=secθandby=tanθ

\textsf{Using}Using

\boxed{\mathsf{sec^2\theta-tan^2\theta=1}}sec2θ−tan2θ=1

\implies\mathsf{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}⟹a2x2−b2y2=1

\implies\mathsf{\frac{b^2x^2-a^2y^2}{a^2b^2}=1}⟹a2b2b2x2−a2y2=1

\implies\mathsf{b^2x^2-a^2y^2=a^2b^2}⟹b2x2−a2y2=a2b2