Question

Locus of the point (sec theta,tan theta)

Answer 1

for (x, y) = $(sec \theta , \: tan \theta)$ ... (i)

we know that,

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

1 + y² = x² ... (from i)

x² - y² - 1 = 0

Hence, locus of the point $(sec \theta , \: tan \theta)$ is x² - y² - 1 = 0