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) $(4\sec{\theta - 1,\ 2\tan{\theta - 2)}}$
D) $(4\sec{\theta - 4,\ 2\tan{\theta - 2)}}$
Explanation
(b)
Any point on 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.