Porisma

Worked solution

Six conics: complete the square and identify graph features

Sketch the following without the use of a graphic calculator, indicating clearly any axial intercepts, asymptotes, vertices/turning points and lines of symmetry.

(a) x2+x−3y=2x^2+x-3y=2

(b) y2+y−3x=2y^2+y-3x=2

(c) a2x2+b2y2=r2a^2x^2+b^2y^2=r^2, a>b>0a>b>0 and r>0r>0

(d) x2+4y2+4x−8y+4=0x^2+4y^2+4x-8y+4=0

(e) 36y2=4x2+136y^2=4x^2+1

(f) x2−4y2−2x−8y−39=0x^2-4y^2-2x-8y-39=0

CJC · Discussion · Question 1

Answer

(a) Upward parabola, vertex (−12,−34)(-\tfrac12,-\tfrac34); (b) rightward parabola, vertex (−34,−12)(-\tfrac34,-\tfrac12); (c) ellipse centred at (0,0)(0,0), semi-axes r/a,r/br/a,r/b; (d) ellipse centred at (−2,1)(-2,1), semi-axes 2,12,1; (e) vertical hyperbola, vertices (0,±16)(0,\pm\tfrac16); (f) horizontal hyperbola, vertices (−5,−1),(7,−1)(-5,-1),(7,-1). Full intercepts, asymptotes and symmetry lines are in the working.

(a)

Complete the square: (x+12)2−14−3y=2(x+\tfrac12)^2-\tfrac14-3y=2, hence y=13(x+12)2−34y=\tfrac13(x+\tfrac12)^2-\tfrac34. This upward parabola has vertex/minimum (−12,−34)(-\tfrac12,-\tfrac34), intercepts (−2,0),(1,0),(0,−23)(-2,0),(1,0),(0,-\tfrac23), symmetry x=−12x=-\tfrac12 and no asymptotes.

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(b)

(y+12)2=3x+94(y+\tfrac12)^2=3x+\tfrac94, so y=−12±3x+94y=-\tfrac12\pm\sqrt{3x+\tfrac94}. The rightward parabola has intercepts (−23,0),(0,1),(0,−2)(-\tfrac23,0),(0,1),(0,-2), symmetry y=−12y=-\tfrac12 and no asymptotes. Its vertex is (−34,−12)(-\tfrac34,-\tfrac12). The source lists turning points as “NIL”: this means no horizontal stationary turning points; the sideways vertex still exists.

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(c)

Divide by r2r^2: x2(r/a)2+y2(r/b)2=1\dfrac{x^2}{(r/a)^2}+\dfrac{y^2}{(r/b)^2}=1. The ellipse is centred at (0,0)(0,0) with intercepts (±r/a,0),(0,±r/b)(\pm r/a,0),(0,\pm r/b), minimum (0,−r/b)(0,-r/b), maximum (0,r/b)(0,r/b), symmetry x=0,y=0x=0,y=0 and no asymptotes. Since a>b>0a>b>0 and r>0r>0, r/a<r/br/a<r/b: the major axis is vertical.

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(d)

Group and complete both squares: (x+2)2−4+4[(y−1)2−1]+4=0(x+2)^2-4+4[(y-1)^2-1]+4=0. Thus (x+2)222+(y−1)212=1\dfrac{(x+2)^2}{2^2}+\dfrac{(y-1)^2}{1^2}=1. Centre (−2,1)(-2,1); intercepts (−2,0),(0,1)(-2,0),(0,1); minimum (−2,0)(-2,0), maximum (−2,2)(-2,2); symmetry x=−2,y=1x=-2,y=1; no asymptotes. The other horizontal endpoint is (−4,1)(-4,1).

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(e)

36y2−4x2=136y^2-4x^2=1, so y2(1/6)2−x2(1/2)2=1\dfrac{y^2}{(1/6)^2}-\dfrac{x^2}{(1/2)^2}=1. This vertical hyperbola has intercepts/vertices (0,±16)(0,\pm\tfrac16), upper-branch minimum (0,16)(0,\tfrac16), lower-branch maximum (0,−16)(0,-\tfrac16) and symmetry x=0,y=0x=0,y=0. Its asymptotes satisfy 36y2−4x2=036y^2-4x^2=0, giving y=±x/3y=\pm x/3.

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(f)

(x−1)2−1−4[(y+1)2−1]−39=0(x-1)^2-1-4[(y+1)^2-1]-39=0, hence (x−1)262−(y+1)232=1\dfrac{(x-1)^2}{6^2}-\dfrac{(y+1)^2}{3^2}=1. Centre (1,−1)(1,-1); vertices (−5,−1),(7,−1)(-5,-1),(7,-1); symmetry x=1,y=−1x=1,y=-1; no horizontal stationary turning points. At y=0y=0, x2−2x−39=0x^2-2x-39=0 gives x=1±210≈−5.32,7.32x=1\pm2\sqrt{10}\approx-5.32,7.32. There is no real yy-intercept. The asymptotes are y+1=±(x−1)/2y+1=\pm(x-1)/2, that is y=x/2−3/2y=x/2-3/2 and y=−x/2−1/2y=-x/2-1/2.

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Compare your sketch

Expand diagram

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Six panels: (a) upward parabola, vertex (−1/2,−3/4), intercepts (−2,0),(1,0),(0,−2/3); (b) rightward parabola, vertex (−3/4,−1/2), intercepts (−2/3,0),(0,1),(0,−2); (c) vertical-major-axis ellipse, intercepts ±r/a horizontally and ±r/b vertically; (d) ellipse centred at (−2,1), endpoints (−4,1),(0,1),(−2,0),(−2,2); (e) hyperbola with vertices (0,±1/6), asymptotes y=±x/3; (f) hyperbola centred at (1,−1), vertices (−5,−1),(7,−1), x-intercepts 1±2√10, asymptotes y=x/2−3/2 and y=−x/2−1/2. Symmetry lines and all features are detailed in each panel and the working.
Six panels: (a) upward parabola, vertex (−1/2,−3/4), intercepts (−2,0),(1,0),(0,−2/3); (b) rightward parabola, vertex (−3/4,−1/2), intercepts (−2/3,0),(0,1),(0,−2); (c) vertical-major-axis ellipse, intercepts ±r/a horizontally and ±r/b vertically; (d) ellipse centred at (−2,1), endpoints (−4,1),(0,1),(−2,0),(−2,2); (e) hyperbola with vertices (0,±1/6), asymptotes y=±x/3; (f) hyperbola centred at (1,−1), vertices (−5,−1),(7,−1), x-intercepts 1±2√10, asymptotes y=x/2−3/2 and y=−x/2−1/2. Symmetry lines and all features are detailed in each panel and the working.
Six panels: (a) upward parabola, vertex (−1/2,−3/4), intercepts (−2,0),(1,0),(0,−2/3); (b) rightward parabola, vertex (−3/4,−1/2), intercepts (−2/3,0),(0,1),(0,−2); (c) vertical-major-axis ellipse, intercepts ±r/a horizontally and ±r/b vertically; (d) ellipse centred at (−2,1), endpoints (−4,1),(0,1),(−2,0),(−2,2); (e) hyperbola with vertices (0,±1/6), asymptotes y=±x/3; (f) hyperbola centred at (1,−1), vertices (−5,−1),(7,−1), x-intercepts 1±2√10, asymptotes y=x/2−3/2 and y=−x/2−1/2. Symmetry lines and all features are detailed in each panel and the working.

Common mistakes

A horizontal vertex can have a vertical tangent. Do not call it a horizontal stationary turning point.

Solution notes

The source’s “vetices” spelling is normalized to “vertices”; its NIL turning-point entries in (b) and (f) refer to horizontal stationary points. Part (c) uses a labelled example r/a=1,r/b=2; other positive ratios with r/a<r/b have the same orientation.