A curve $C_1$ has equation $(ky)^2-8ky-x^2+2x=-11$, where $k0$. Show that the equatio

Question

A curve $C_1$ has equation $(ky)^2-8ky-x^2+2x=-11$, where $k>0$. Show that the equation of $C_1$ can be expressed as $\alpha\left(y-\frac{4}{k}\right)^2-\beta(x-1)^2=1$, where $\alpha$ is a constant in terms of $k$ and $\beta$ is a constant to be determined.[3] It is given that one of the asymptotes of $C_1$ has equation $y=\frac{1}{2}x+\frac{3}{2}$. Find the value of $k$. Hence write down the equation of the other asymptote.[3] Using the value of $k$ found in part (b), sketch $C_1$, stating clearly the equations of its asymptotes and the coordinates of any turning points. [3]

Accepted Answer

(a) $\alpha=\dfrac{k^2}{4}, \beta=\dfrac14$ (b) $k=2$, other asymptote $y=-\dfrac x2+\dfrac52$ (c) centre $(1,2)$, turning points $(1,1),(1,3)$

2025 HCI Promo Q4

2025 HCI Promo Q4

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