The curve ${{C}_{1}}$ has equation ${{x}^{2}}-{{y}^{2}}+6y+16=0$ The curve ${{C}_{2}}$ has equ

Question

The curve ${{C}_{1}}$ has equation ${{x}^{2}}-{{y}^{2}}+6y+16=0$ The curve ${{C}_{2}}$ has equation ${{\left( \frac{x-p}{5} \right)}^{2}}-{{\left( \frac{y-q}{5} \right)}^{2}}=1$, where $p$ and $q$ are constants. Sketch ${{C}_{1}}$, stating the coordinates of any points of intersection with the $x$- and $y$-axes and the equations of any asymptotes.[3] The line $y=mx+c$ intersects ${{C}_{1}}$ twice. State the range of values of $m$.[2] Given that ${{C}_{1}}$ and ${{C}_{2}}$ have the same asymptotes, determine the values of $p$ and $q$.[2] On the same diagram in part (i), sketch ${{C}_{2}}$.[1] Let $n$ be the number of points of intersection of the curve ${{x}^{2}}+{{(y-3)}^{2}}={{r}^{2}}$ where $r>0$, with the curve ${{C}_{2}}$. Determine the possible values of $n$.[1]

Accepted Answer

(i) $C_1:\dfrac{(y-3)^2}{25}-\dfrac{x^2}{25}=1$, $y$-intercepts $(0,8)$ and $(0,-2)$, no $x$-intercepts, asymptotes $y=3\pm x$ (ii) As written, the range is underdetermined because it depends on $c$. For $m\ne\pm1$, two intersections require $(c-3)^2+25(m^2-1)>0$ (iii) $p=0$, $q=3$ (iv) $C_2:\dfrac{x^2}{25}-\dfrac{(y-3)^2}{25}=1$, vertices $(\pm5,3)$ (v) $n=0,2\text{ or }4$

2014 JJC Promo Q6

2014 JJC Promo Q6

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