Question

The locus of represented by $x=\frac{a}{2}(t+\frac{1}{t}),y=\frac{a}{2}(t-\frac{1}{t})$ is:

• A. $x^2+y^2=a^2$
• B. $x^2-y^2=a^2$
• C. $2x^2-y^2=a^2$
• D. $x^2-2y^2=a^2$

Answer 1

The correct option is B $x^2-y^2=a^2$

$x=\frac{a}{2}(t+\frac{1}{t})$
Hence
$x^2=\frac{a^2}{4}(t^2+\frac{1}{t^2}+2)$ ...$\left(i\right)$
And
$y=\frac{a}{2}(t-\frac{1}{t})$
Hence
$y^2=\frac{a^2}{4}(t^2+\frac{1}{t^2}-2)$....$\left(ii\right)$
Subtracting $\left(ii\right)$ from $\left(i\right)$ gives us
$x^2-y^2=\frac{a^2}{4}\left(4\right)$
$x^2-y^2=a^2$