Conic Sections and Parametric Equations: practice solutions, Grade 12 – download the PDF

1 Center and radius ★★★

Compare with \((x-h)^2+(y-k)^2=r^2\): \(h=4\), \(k=-7\) (because \(y+7=y-(-7)\)) and \(r^2=49\).

The center is \((4,-7)\) and the radius is \(7\).

2 Writing a circle equation ★★★

\(h=-2\), \(k=5\), \(r=3\), so \((x-(-2))^2+(y-5)^2=3^2\).

\((x+2)^2+(y-5)^2=9\).

3 A simple parabola ★★★

The form is \(x^2=4py\), so \(4p=8\) and \(p=2\).

Vertex \((0,0)\), focus \((0,2)\), directrix \(y=-2\).

4 Vertices of an ellipse ★★★

\(a^2=49\), so \(a=7\); \(b^2=25\), so \(b=5\). The long axis is horizontal.

Vertices \((\pm7,0)\), co-vertices \((0,\pm5)\). \(c^2=49-25=24\), so \(c=\sqrt{24}=2\sqrt6\approx4.90\).

5 Hyperbola features ★★★

\(a=3\), \(b=4\), \(c^2=9+16=25\), so \(c=5\).

Vertices \((\pm3,0)\), foci \((\pm5,0)\), asymptotes \(y=\pm\dfrac{4}{3}x\).

6 Name that conic ★★★

1. Sum of squares with equal coefficients: a circle of radius \(6\).
2. Dividing by \(36\): \(\dfrac{x^2}{9}+\dfrac{y^2}{4}=1\), an ellipse.
3. Difference of squares equal to \(1\): a hyperbola opening up and down.
4. Only one variable is squared: a parabola (\(x^2=\dfrac{1}{2}y\), so \(p=\dfrac18\)).

7 A table of points ★★★

\(t=0\): \((2,-1)\); \(t=1\): \((3,2)\); \(t=2\): \((4,5)\); \(t=3\): \((5,8)\).

From \(x=t+2\), \(t=x-2\). Then \(y=3(x-2)-1=3x-7\). The curve is the line \(y=3x-7\).

8 Completing the square ★★★

Group: \((x^2+8x)+(y^2-6y)=-9\). Add \(16\) and \(9\) to both sides: \((x+4)^2+(y-3)^2=-9+16+9=16\).

Center \((-4,3)\), radius \(4\).

9 Circle through a point ★★★

The radius is the distance: \(r=\sqrt{(4-1)^2+(6-2)^2}=\sqrt{25}=5\).

Standard form: \((x-1)^2+(y-2)^2=25\). Expanding: \(x^2-2x+1+y^2-4y+4=25\), so the general form is \(x^2+y^2-2x-4y-20=0\).

10 From focus and directrix ★★★

The vertex is halfway between them: \((0,0)\), and \(p=5\). So \(x^2=4\cdot5\,y\), that is \(x^2=20y\).

Check: \(10^2=100=20\cdot5\). The distance to the focus is \(10\) and the distance to the directrix is \(5-(-5)=10\).

11 Satellite dish ★★★

The equation is \(x^2=4py\). The rim point is \((5,2)\) (half the width, full depth): \(25=4p\cdot2=8p\).

So \(p=\dfrac{25}{8}=3.125\). The receiver is 3.125 ft (about 0.95 m) above the vertex.

12 Ellipse from foci and vertices ★★★

\(a=5\), \(c=3\), so \(b^2=a^2-c^2=25-9=16\).

\(\dfrac{x^2}{25}+\dfrac{y^2}{16}=1\) and \(e=\dfrac{c}{a}=\dfrac{3}{5}=0.6\).

13 Hyperbola from foci and vertices ★★★

\(a=6\), \(c=10\), so \(b^2=c^2-a^2=100-36=64\) and \(b=8\).

Equation: \(\dfrac{x^2}{36}-\dfrac{y^2}{64}=1\). Asymptotes: \(y=\pm\dfrac{8}{6}x=\pm\dfrac{4}{3}x\). Eccentricity: \(e=\dfrac{10}{6}=\dfrac{5}{3}\approx1.67\).

14 Eliminating by substitution ★★★

\(t=x+3\). Then \(y=(x+3)^2+2(x+3)=x^2+6x+9+2x+6=x^2+8x+15\).

It is the parabola \(y=x^2+8x+15\), opening upward.

15 Trigonometric elimination ★★★

\(\cos t=\dfrac{x}{4}\), \(\sin t=\dfrac{y}{3}\). Using \(\cos^2t+\sin^2t=1\): \(\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\).

An ellipse with \(a=4\), \(b=3\), traced once counterclockwise starting at \((4,0)\).

16 Find the eccentricity first ★★★

\(c=ea=0.6\cdot10=6\). Then \(b^2=a^2-c^2=100-36=64\), so \(b=8\).

\(\dfrac{x^2}{100}+\dfrac{y^2}{64}=1\).

17 An ellipse in general form ★★★

\(4(x^2-4x)+9(y^2+2y)=11\). Complete the squares: \(4(x-2)^2-16+9(y+1)^2-9=11\), so \(4(x-2)^2+9(y+1)^2=36\).

Divide by \(36\): \(\dfrac{(x-2)^2}{9}+\dfrac{(y+1)^2}{4}=1\). Center \((2,-1)\), \(a=3\), \(b=2\), \(c=\sqrt5\).

Foci \((2\pm\sqrt5,-1)\); \(e=\dfrac{\sqrt5}{3}\approx0.75\).

18 A vertical hyperbola ★★★

\(9(y^2-4y)-4(x^2+2x)=4\). Complete the squares: \(9(y-2)^2-36-4(x+1)^2+4=4\), so \(9(y-2)^2-4(x+1)^2=36\).

Standard form: \(\dfrac{(y-2)^2}{4}-\dfrac{(x+1)^2}{9}=1\). Center \((-1,2)\), \(a=2\), \(b=3\), \(c=\sqrt{13}\).

Vertices \((-1,0)\) and \((-1,4)\); foci \((-1,2\pm\sqrt{13})\); asymptotes \(y-2=\pm\dfrac{2}{3}(x+1)\).

19 A sideways parabola ★★★

\(y^2-6y=8x-25\). Add \(9\): \((y-3)^2=8x-16=8(x-2)\).

So \(4p=8\), \(p=2\), vertex \((2,3)\). Focus \((4,3)\), directrix \(x=0\).

20 A planet’s orbit ★★★

The two extreme distances add up to the major axis: \(2a=90+150=240\), so \(a=120\). The star is at distance \(c\) from the center, and the closest distance is \(a-c=90\), so \(c=30\).

\(e=\dfrac{30}{120}=0.25\). \(b=\sqrt{120^2-30^2}=\sqrt{13\,500}=30\sqrt{15}\approx116.2\).

\(a=120\), \(c=30\), \(e=0.25\), \(b\approx116.2\) million miles.

21 Line and circle ★★★

Substitute: \(x^2+(x+1)^2=25\), so \(2x^2+2x-24=0\), that is \(x^2+x-12=0\), or \((x+4)(x-3)=0\).

\(x=3\Rightarrow y=4\); \(x=-4\Rightarrow y=-3\). The points are \((3,4)\) and \((-4,-3)\). Both satisfy \(x^2+y^2=25\).

22 Projectile path ★★★

1. \(t=\dfrac{x}{30}\), so \(y=5+\dfrac{4}{3}x-\dfrac{4}{225}x^2\), a downward parabola.
2. The maximum is at \(t=\dfrac{40}{32}=1.25\) s: \(y=5+50-25=30\) ft (about 9.1 m).
3. Solve \(y=0\): \(16t^2-40t-5=0\), \(t=\dfrac{40+\sqrt{1920}}{32}\approx2.619\) s. Then \(x=30t\approx78.6\) ft (about 24.0 m).

23 Only part of a curve ★★★

\(\sin^2t=1-\cos^2t\), so \(y=1-x^2\).

But \(x=\cos t\) only takes values in \([-1,1]\), and \(y\) in \([0,1]\). Only the arc of the parabola \(y=1-x^2\) between \((-1,0)\) and \((1,0)\) is drawn, traced back and forth as \(t\) increases.

24 A hyperbola with sec and tan ★★★

\(\dfrac{x}{2}=\sec t\) and \(\dfrac{y}{3}=\tan t\). Since \(\sec^2t-\tan^2t=1\): \(\dfrac{x^2}{4}-\dfrac{y^2}{9}=1\).

Here \(a=2\), \(b=3\), so the asymptotes are \(y=\pm\dfrac{3}{2}x\).