The curve $C$ has equation $\frac{(y-3)^2}{5}-\frac{(x+2)^2}{9}=1$. Sketch $C$, stating the eq

Question

The curve $C$ has equation $\frac{(y-3)^2}{5}-\frac{(x+2)^2}{9}=1$. Sketch $C$, stating the equations of the asymptotes and their point of intersection.[3] Hence state the exact range of values of $m$ such that the straight line $y=m(x+2)+3$ intersect $C$.[2]

Accepted Answer

(i) Vertices $(-2,3\pm\sqrt5)$; asymptotes $y-3=\pm\frac{\sqrt5}{3}(x+2)$, intersecting at $(-2,3)$ (ii) $m\frac{\sqrt5}{3}$

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