Porisma

Worked solution

Question 1

Sketch the curve with equation 16y2−16y−4x2=−316y^2-16y-4x^2=-3, indicating clearly the equation of any asymptote (s) and the coordinates of any stationary point (s). marks

2012/HCI/Promo/12(a)

Answer

The curve is a hyperbola with centre (0,12)\left(0,\tfrac12\right) and asymptotes y=12+12xy=\tfrac12+\tfrac12x and y=12−12xy=\tfrac12-\tfrac12x. Its stationary points are (0,14)\left(0,\tfrac14\right) and (0,34)\left(0,\tfrac34\right). The two branches open upwards and downwards from these points.

Compare your reasoning

Complete the square in yy:

16(y−12)2−4−4x2=−3⟹(y−12)2(14)2−x2(12)2=1.16\left(y-\tfrac12\right)^2-4-4x^2=-3\quad\Longrightarrow\quad \frac{\left(y-\frac12\right)^2}{\left(\frac14\right)^2}-\frac{x^2}{\left(\frac12\right)^2}=1.

The asymptotes have gradient ±1/41/2=±12\pm\dfrac{1/4}{1/2}=\pm\dfrac12 through the centre. The vertices are 14\tfrac14 above and below the centre.

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Compare your sketch

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Hyperbola centred at (0, 1/2), with upper vertex (0, 3/4) and lower vertex (0, 1/4). Dashed asymptotes are y = 1/2 ± x/2. Both branches approach the asymptotes.
Hyperbola centred at (0, 1/2), with upper vertex (0, 3/4) and lower vertex (0, 1/4). Dashed asymptotes are y = 1/2 ± x/2. Both branches approach the asymptotes.
Hyperbola centred at (0, 1/2), with upper vertex (0, 3/4) and lower vertex (0, 1/4). Dashed asymptotes are y = 1/2 ± x/2. Both branches approach the asymptotes.