Q. Any point on the hyperbola $\frac{(x + 1)^{2}}{16} - \frac{(y - 2)^{2}}{4} = 1$ is of the form
A) $(4\ sec\ \theta,\ 2\tan\theta$)
B) $(4\ sec\ \theta + 1,\ 2\tan\theta - 2)$
C) $(4\ sec\ \theta - 1,\ 2tan\theta - 2)$
D) $(4\ sec\ \theta - 1,\ 2\tan{\theta + 2)}$
Explanation
(d)
Any point on the hyperbola
$\frac{(x + 1)^{2}}{16} - \frac{(y - 2)^{2}}{4} = 1,\$is of the form
$(4\sec{\theta - 1,\ 2\tan{\theta + 2)}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.