Conic sections 2 Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Conic sections 2 questions. See exactly how to solve problems on conics, eccentricity, vertices, hyperbola.

conicseccentricityverticeshyperbolaasymptotesellipse
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The ellipse CC has equation x225+y29=1\frac{x^{2}}{25}+\frac{y^{2}}{9}=1. Find the eccentricity of CC.

Worked solution

  1. Identify a2a^{2} and b2b^{2} from the equation

    a2=25,b2=9a^{2}=25,\quad b^{2}=9

    The equation is already in the standard form for an ellipse.

  2. Quote the eccentricity relation for an ellipse

    b2=a2(1−e2)b^{2}=a^{2}\left(1-e^{2}\right)

    This is the relation given in the formula book.

  3. State the eccentricity of CC

    e=45e=\frac{4}{5}

    This is the eccentricity of the ellipse.

Answer
45\frac{4}{5}
Question 2
2 markseasy
The ellipse CC has equation x225+y216=1\frac{x^{2}}{25}+\frac{y^{2}}{16}=1. Find the eccentricity of CC.

Worked solution

  1. Identify a2a^{2} and b2b^{2} from the equation

    a2=25,b2=16a^{2}=25,\quad b^{2}=16

    The equation is already in the standard form for an ellipse.

  2. Quote the eccentricity relation for an ellipse

    b2=a2(1−e2)b^{2}=a^{2}\left(1-e^{2}\right)

    This is the relation given in the formula book.

  3. Substitute the values of a2a^{2} and b2b^{2}

    16=25(1−e2)16=25\left(1-e^{2}\right)

    Both a2a^{2} and b2b^{2} are read straight off the equation.

  4. State the eccentricity of CC

    e=35e=\frac{3}{5}

    This is the eccentricity of the ellipse.

Answer
35\frac{3}{5}
Question 3
2 markseasy
The hyperbola CC has equation x216−y29=1\frac{x^{2}}{16}-\frac{y^{2}}{9}=1. Find the eccentricity of CC.

Worked solution

  1. Identify a2a^{2} and b2b^{2} from the equation

    a2=16,b2=9a^{2}=16,\quad b^{2}=9

    The equation is already in the standard form for a hyperbola.

  2. Quote the eccentricity relation for a hyperbola

    b2=a2(e2−1)b^{2}=a^{2}\left(e^{2}-1\right)

    This is the relation given in the formula book.

  3. Substitute the values of a2a^{2} and b2b^{2}

    9=16(e2−1)9=16\left(e^{2}-1\right)

    Both a2a^{2} and b2b^{2} are read straight off the equation.

  4. State the eccentricity of CC

    e=54e=\frac{5}{4}

    This is the eccentricity of the hyperbola.

Answer
54\frac{5}{4}
Question 4
2 markseasy
The hyperbola CC has equation x29−y216=1\frac{x^{2}}{9}-\frac{y^{2}}{16}=1. Find the eccentricity of CC.

Worked solution

  1. Identify a2a^{2} and b2b^{2} from the equation

    a2=9,b2=16a^{2}=9,\quad b^{2}=16

    The equation is already in the standard form for a hyperbola.

  2. Quote the eccentricity relation for a hyperbola

    b2=a2(e2−1)b^{2}=a^{2}\left(e^{2}-1\right)

    This is the relation given in the formula book.

  3. State the eccentricity of CC

    e=53e=\frac{5}{3}

    This is the eccentricity of the hyperbola.

Answer
53\frac{5}{3}
Question 5
2 markseasy
The ellipse CC has equation x225+y29=1\frac{x^{2}}{25}+\frac{y^{2}}{9}=1. Find the coordinates of the points where CC crosses the xx-axis.

Worked solution

  1. Identify a2a^{2} and b2b^{2} from the equation

    a2=25,b2=9a^{2}=25,\quad b^{2}=9

    The equation is already in the standard form for an ellipse.

  2. Set y=0y=0 in the equation of CC

    x225=1\frac{x^{2}}{25}=1

    The curve meets the xx-axis where y=0y=0.

  3. Solve for x2x^{2}

    x2=25x^{2}=25

    Multiply both sides by a2a^{2}.

  4. State the coordinates

    (±5,0)\left(\pm 5,0\right)

    These are the vertices of the ellipse on the xx-axis.

Answer
(±5,0)\left(\pm 5,0\right)

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