Question

Find the locus of the point x=a+b sec teta,y=b+a tan teta

Answer 1

$\textbf{Given:}$

$x=a+b\,sec\theta$

$y=b+a\,tan\theta$

$\textbf{To find:}$

$\text{The locus of (x,y)}$

$\textbf{Solution:}$

$\text{Consider,}$

$x=a+b\,sec\theta$

$\implies\,sec\theta=\dfrac{x-a}{b}$

$y=b+a\,tan\theta$

$\implies\,tan\theta=\dfrac{y-b}{a}$

$\text{We know that,}$

$\bf\,sec^2\theta-tan^2\theta=1$

$\implies\,(\dfrac{x-a}{b})^2-(\dfrac{y-b}{a})^2=1$

$\implies\,\dfrac{(x-a)^2}{b^2}-\dfrac{(y-b)^2}{a^2}=1$

$\implies\,\dfrac{a^2(x-a)^2-b^2(y-b)^2}{a^2b^2}=1$

$\implies\bf\,a^2(x-a)^2-b^2(y-b)^2=a^2b^2$

$\therefore\textbf{The locus of (x,y) is}$

$\boxed{\bf\,a^2(x-a)^2-b^2(y-b)^2=a^2b^2}$

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