Question

Any point on the hyperbola $\frac{(x+1)^2}{16}-\frac{(y-2)^2}{4}=1$ is of the form

(A) $(4 \sec \theta, 2 \tan e)$

(B) $(4 \sec \theta+1,2 \tan \theta-2)$

(C) $(4 \sec \theta-1,2 \tan \theta-2)$

(D) $(4 \sec \theta-1,2 \tan \theta+2)$

Answer 1

Any point on hyperbola $\frac{(x+1)^2}{16}-\frac{(y-2)^2}{4}=1$ is of the form $(4 \sec \theta-1,2 \tan \theta+2)$

So, The correct option will be (D).