The curve with equation $\frac{{{y}^{2}}}{9}-{{x}^{2}}=1$ undergoes a two-step transformations to

Question

The curve with equation $\frac{{{y}^{2}}}{9}-{{x}^{2}}=1$ undergoes a two-step transformations to become curve $C$ with equation $\frac{{{y}^{2}}}{9}-\frac{{{\left( x-1 \right)}^{2}}}{4}=1$. State the two transformations involved for the curve $\frac{{{y}^{2}}}{9}-{{x}^{2}}=1$.[2] Draw a sketch of the curve $C$, labelling clearly the equation(s) of its asymptote(s), intersection with the axes and the coordinates of any turning points.[3] Show that the point $\left( -1,-3 \right)$ lies on the line $y=mx+m-3$ for all real values of $m$.[1] Hence using the diagram drawn in (ii), find the range of values of $k$ such that the equation $\frac{{{\left( kx+k-3 \right)}^{2}}}{9}-\frac{{{\left( x-1 \right)}^{2}}}{4}=1$ has ${2}$ negative real roots.[2]

Accepted Answer

(i) Stretch by scale factor $2$ parallel to the $x$-axis, then translate $1$ unit in the positive $x$-direction (ii) Centre $(1,0)$, turning points $(1,3)$ and $(1,-3)$, asymptotes $y=\dfrac32(x-1)$ and $y=-\dfrac32(x-1)$, $y$-intercepts $\left(0,\dfrac{3\sqrt5}{2}\right)$ and $\left(0,-\dfrac{3\sqrt5}{2}\right)$, no $x$-intercepts (iii) $m(-1)+m-3=-3$, so $(-1,-3)$ lies on every line (iv) $k3+\dfrac{3\sqrt5}{2}$

2018 SRJC P1 Q4

2018 SRJC P1 Q4

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