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Worked solution

Two horizontal hyperbolas: algebra and calculator checks

Sketch the hyperbolas:

(a) x24−y2=1\dfrac{x^2}{4}-y^2=1,

(b) (x+4)29−(y−2)216=1\dfrac{(x+4)^2}{9}-\dfrac{(y-2)^2}{16}=1,

and state the line(s) of symmetry.

CJC · Lecture · Example 5

(a) Algebraic method

(x−0)222−(y−0)212=1\dfrac{(x-0)^2}{2^2}-\dfrac{(y-0)^2}{1^2}=1, so the centre is (0,0)(0,0) and the vertices are (−2,0),(2,0)(-2,0),(2,0). Set the right side to zero for asymptotes: x2/4−y2=0x^2/4-y^2=0, hence y2=x2/4y^2=x^2/4 and y=±x/2y=\pm x/2. The lines of symmetry are x=0,y=0x=0,y=0. Draw two horizontal branches approaching these lines. This working fills the Student copy’s blanks using the compared Teacher copy.

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(b) Algebraic method

[x−(−4)]232−(y−2)242=1\dfrac{[x-(-4)]^2}{3^2}-\dfrac{(y-2)^2}{4^2}=1, giving centre (−4,2)(-4,2) and vertices (−7,2),(−1,2)(-7,2),(-1,2). The homogeneous equation is (x+4)2/9−(y−2)2/16=0(x+4)^2/9-(y-2)^2/16=0, or 16(x+4)2=9(y−2)216(x+4)^2=9(y-2)^2. Thus y−2=±43(x+4)y-2=\pm\tfrac43(x+4) and the asymptotes are y=43x+223y=\tfrac43x+\tfrac{22}{3} and y=−43x−103y=-\tfrac43x-\tfrac{10}{3}. Symmetry lines: x=−4,y=2x=-4,y=2.

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(a) Conics application and window controls

Follow the source table: apps → [2:Conics] → [3:HYPERBOLA]. Press mode for CONIC SETTINGS, choose FUNC (functions) and MAN (manual window settings), then ESC to return. Select [1:], the horizontal form (X−H)2/A2−(Y−K)2/B2=1(X-H)^2/A^2-(Y-K)^2/B^2=1. Enter A=2, B=1, H=0, K=0. To SOLVE, press alpha, enter. The feature screen shows centre (0,0)(0,0), vertices (−2,0),(2,0)(-2,0),(2,0), focus readouts (−2.236,0)(-2.236,0) and (2.2361,0)(2.2361,0), and slope ±0.5\pm0.5. The focus readouts are displayed decimals; the exact focus coordinates are (±5,0)(\pm\sqrt5,0) (an authored check). The centre/vertices give symmetry x=0,y=0x=0,y=0, and the slope gives asymptotes y=±x/2y=\pm x/2. Press graph; trace coordinates may be inaccurate. Press zoom for the displayed choices 1:Zoom Conic, 2:Zoom In, 3:Zoom Out, 4:ZBox, 5:ZSquare, 6:ZStandard. Without MAN, the source says a Zoom Conic command is performed by default. In MAN, press window for manual controls; the screenshot has Xmin=−10, Xmax=10, Xscl=1, Ymin=−10, Ymax=10, Yscl=1. These are source display settings, not part of the equation.

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(b) Conics application

Select horizontal form [1:] and enter A=3, B=4, H=−4, K=2. Press alpha, enter to SOLVE. The feature screen shows centre (−4,2)(-4,2), vertices (−7,2),(−1,2)(-7,2),(-1,2), foci (−9,2),(1,2)(-9,2),(1,2) and slope readout ±1.3333\pm1.3333. The exact slopes are ±4/3\pm4/3. The centre/vertices give symmetry x=−4,y=2x=-4,y=2; combine the slopes with the centre to obtain y−2=±43(x+4)y-2=\pm\tfrac43(x+4), not lines through the origin. Press graph and trace if needed, retaining the source warning about numerical trace accuracy. The exact foci also follow from 32+42=5\sqrt{3^2+4^2}=5 (an authored verification).

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Compare your sketch

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Two smooth two-branch hyperbolas. Part (a): centre(0,0), vertices (−2,0),(2,0), asymptotes y=±x/2, symmetry x=0,y=0. Part (b): centre(−4,2), vertices (−7,2),(−1,2), asymptotes y=(4/3)x+22/3 and y=−(4/3)x−10/3, symmetry x=−4,y=2. Centre points are guides, not points on the curves.
Two smooth two-branch hyperbolas. Part (a): centre(0,0), vertices (−2,0),(2,0), asymptotes y=±x/2, symmetry x=0,y=0. Part (b): centre(−4,2), vertices (−7,2),(−1,2), asymptotes y=(4/3)x+22/3 and y=−(4/3)x−10/3, symmetry x=−4,y=2. Centre points are guides, not points on the curves.
Two smooth two-branch hyperbolas. Part (a): centre(0,0), vertices (−2,0),(2,0), asymptotes y=±x/2, symmetry x=0,y=0. Part (b): centre(−4,2), vertices (−7,2),(−1,2), asymptotes y=(4/3)x+22/3 and y=−(4/3)x−10/3, symmetry x=−4,y=2. Centre points are guides, not points on the curves.

Common mistakes

For a translated hyperbola, asymptote slopes alone are not equations. Use the centre to determine the intercepts.

Solution notes

Student/Teacher PDF pages 17–19, printed pages 129–131. No marks printed. Both source asks, algebraic method, Conics screens, settings and feature values compared. Teacher fills the first part’s working and symmetry answers.

The Student sketch in (a) places the two vertex dots above the x-axis despite labelling them (−2,0),(2,0). The equation and Teacher sketch put them on the axis; the reconstruction follows those consistent coordinates and records the original discrepancy here.

Source screenshots retained in the original PDFs; mathematical input, feature readouts, menu/window settings and warnings transcribed into accessible working. Decimal focus/slope readouts are not substituted for exact values.

Read the hint