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NCERT Solutions · Class 11 Mathematics Conic Sections

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EXERCISE 10.4 1–10 (part 6 of 8)

  1. In each of the Exercises $\displaystyle 1$ to $\displaystyle 6$, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

    Exercise 1

    x216−y29=1\displaystyle \frac{x^{2}}{16}-\frac{y^{2}}{9}=1

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    NCERT’s answer
    Foci $\displaystyle ( \pm 5,0)$, Vertices $\displaystyle ( \pm 4,0) ; e=\frac{5}{4}$; Latus rectum $\displaystyle =\frac{9}{2}$
    Comparing with the standard hyperbola. The equation \(\displaystyle \dfrac{x^{2}}{16}-\dfrac{y^{2}}{9}=1\) has the \(\displaystyle x^{2}\) term positive, so it matches \[\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1,\] whose transverse axis is the \(\displaystyle x\)-axis. Hence \[a^{2}=16,\quad b^{2}=9\ \Rightarrow\ a=4,\quad b=3.\] For a hyperbola \(\displaystyle c^{2}=a^{2}+b^{2}\) (note the plus sign, unlike the ellipse): \[c^{2}=16+9=25\ \Rightarrow\ c=5.\] Now read off the four required quantities.
    Foci \(\displaystyle (\pm c,0)=(\pm 5,0)\)
    Vertices \(\displaystyle (\pm a,0)=(\pm 4,0)\)
    Eccentricity \(\displaystyle e=\dfrac{c}{a}=\dfrac{5}{4}\)
    Latus rectum \(\displaystyle =\dfrac{2b^{2}}{a}=\dfrac{2(9)}{4}=\dfrac{9}{2}\)
    Foci \(\displaystyle (\pm 5,0)\); vertices \(\displaystyle (\pm 4,0)\); \(\displaystyle e=\dfrac{5}{4}\); latus rectum \(\displaystyle =\dfrac{9}{2}\)

  2. Exercise 2

    y29−x227=1\displaystyle \frac{y^{2}}{9}-\frac{x^{2}}{27}=1

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    NCERT’s answer
    Foci $\displaystyle (0 \pm 6)$, Vertices $\displaystyle (0, \pm 3) ; e=2$; Latus rectum $\displaystyle =18$
    Comparing with the standard hyperbola. Here the \(\displaystyle y^{2}\) term is positive, so the equation matches \[\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1,\] whose transverse axis is the \(\displaystyle y\)-axis. Hence \[a^{2}=9,\quad b^{2}=27\ \Rightarrow\ a=3,\quad b=3\sqrt{3}.\] Using \(\displaystyle c^{2}=a^{2}+b^{2}\), \[c^{2}=9+27=36\ \Rightarrow\ c=6.\]
    Foci \(\displaystyle (0,\pm c)=(0,\pm 6)\)
    Vertices \(\displaystyle (0,\pm a)=(0,\pm 3)\)
    Eccentricity \(\displaystyle e=\dfrac{c}{a}=\dfrac{6}{3}=2\)
    Latus rectum \(\displaystyle =\dfrac{2b^{2}}{a}=\dfrac{2(27)}{3}=18\)
    Foci \(\displaystyle (0,\pm 6)\); vertices \(\displaystyle (0,\pm 3)\); \(\displaystyle e=2\); latus rectum \(\displaystyle =18\)

  3. Exercise 3

    9y2−4x2=36\displaystyle 9 y^{2}-4 x^{2}=36

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    NCERT’s answer
    Foci $\displaystyle (0, \pm \sqrt{13})$, Vertices $\displaystyle (0, \pm 2) ; e=\frac{\sqrt{13}}{2}$; Latus rectum = $\displaystyle 9$
    Reduce to standard form first. Divide \(\displaystyle 9y^{2}-4x^{2}=36\) throughout by \(\displaystyle 36\): \[\frac{y^{2}}{4}-\frac{x^{2}}{9}=1.\] The \(\displaystyle y^{2}\) term is positive, so this is \(\displaystyle \dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1\) with transverse axis along the \(\displaystyle y\)-axis, and \[a^{2}=4,\quad b^{2}=9\ \Rightarrow\ a=2,\quad b=3.\] Using \(\displaystyle c^{2}=a^{2}+b^{2}\), \[c^{2}=4+9=13\ \Rightarrow\ c=\sqrt{13}.\]
    Foci \(\displaystyle (0,\pm\sqrt{13})\)
    Vertices \(\displaystyle (0,\pm 2)\)
    Eccentricity \(\displaystyle e=\dfrac{c}{a}=\dfrac{\sqrt{13}}{2}\)
    Latus rectum \(\displaystyle =\dfrac{2b^{2}}{a}=\dfrac{2(9)}{2}=9\)
    Foci \(\displaystyle (0,\pm\sqrt{13})\); vertices \(\displaystyle (0,\pm 2)\); \(\displaystyle e=\dfrac{\sqrt{13}}{2}\); latus rectum \(\displaystyle =9\)

  4. Exercise 4

    16x2−9y2=576\displaystyle 16 x^{2}-9 y^{2}=576

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    NCERT’s answer
    Foci $\displaystyle ( \pm 10,0)$, Vertices $\displaystyle ( \pm 6,0) ; e=\frac{5}{3}$; Latus rectum $\displaystyle =\frac{64}{3}$
    Reduce to standard form first. Divide \(\displaystyle 16x^{2}-9y^{2}=576\) throughout by \(\displaystyle 576\): \[\frac{x^{2}}{36}-\frac{y^{2}}{64}=1,\] since \(\displaystyle \dfrac{576}{16}=36\) and \(\displaystyle \dfrac{576}{9}=64\). The \(\displaystyle x^{2}\) term is positive, so the transverse axis is the \(\displaystyle x\)-axis and \[a^{2}=36,\quad b^{2}=64\ \Rightarrow\ a=6,\quad b=8.\] Using \(\displaystyle c^{2}=a^{2}+b^{2}\), \[c^{2}=36+64=100\ \Rightarrow\ c=10.\]
    Foci \(\displaystyle (\pm 10,0)\)
    Vertices \(\displaystyle (\pm 6,0)\)
    Eccentricity \(\displaystyle e=\dfrac{c}{a}=\dfrac{10}{6}=\dfrac{5}{3}\)
    Latus rectum \(\displaystyle =\dfrac{2b^{2}}{a}=\dfrac{2(64)}{6}=\dfrac{64}{3}\)
    Foci \(\displaystyle (\pm 10,0)\); vertices \(\displaystyle (\pm 6,0)\); \(\displaystyle e=\dfrac{5}{3}\); latus rectum \(\displaystyle =\dfrac{64}{3}\)

  5. Exercise 5

    5y2−9x2=36\displaystyle 5 y^{2}-9 x^{2}=36

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    NCERT’s answer
    Foci $\displaystyle \left(0, \pm \frac{2 \sqrt{14}}{\sqrt{5}}\right)$, Vertices $\displaystyle \left(0, \pm \frac{6}{\sqrt{5}}\right) ; e=\frac{\sqrt{14}}{3} ;$ Latus rectum $\displaystyle =\frac{4 \sqrt{5}}{3}$
    Reduce to standard form first. Divide \(\displaystyle 5y^{2}-9x^{2}=36\) throughout by \(\displaystyle 36\): \[\frac{y^{2}}{36/5}-\frac{x^{2}}{4}=1.\] The \(\displaystyle y^{2}\) term is positive, so the transverse axis is the \(\displaystyle y\)-axis and \[a^{2}=\frac{36}{5},\quad b^{2}=4\ \Rightarrow\ a=\frac{6}{\sqrt{5}},\quad b=2.\] Using \(\displaystyle c^{2}=a^{2}+b^{2}\), \[c^{2}=\frac{36}{5}+4=\frac{56}{5}\ \Rightarrow\ c=\sqrt{\frac{56}{5}}=\frac{2\sqrt{14}}{\sqrt{5}}=\frac{2\sqrt{70}}{5}.\]
    Foci \(\displaystyle \left(0,\pm\dfrac{2\sqrt{70}}{5}\right)\)
    Vertices \(\displaystyle \left(0,\pm\dfrac{6}{\sqrt{5}}\right)=\left(0,\pm\dfrac{6\sqrt{5}}{5}\right)\)
    Eccentricity \(\displaystyle e=\dfrac{c}{a}=\dfrac{2\sqrt{14}/\sqrt{5}}{6/\sqrt{5}}=\dfrac{2\sqrt{14}}{6}=\dfrac{\sqrt{14}}{3}\)
    Latus rectum \(\displaystyle =\dfrac{2b^{2}}{a}=\dfrac{2(4)}{6/\sqrt{5}}=\dfrac{8\sqrt{5}}{6}=\dfrac{4\sqrt{5}}{3}\)
    Foci \(\displaystyle \left(0,\pm\dfrac{2\sqrt{70}}{5}\right)\); vertices \(\displaystyle \left(0,\pm\dfrac{6\sqrt{5}}{5}\right)\); \(\displaystyle e=\dfrac{\sqrt{14}}{3}\); latus rectum \(\displaystyle =\dfrac{4\sqrt{5}}{3}\)

  6. Exercise 6

    49y2−16x2=784\displaystyle 49 y^{2}-16 x^{2}=784.

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    NCERT’s answer
    Foci $\displaystyle (0, \pm \sqrt{65})$, Vertices $\displaystyle (0, \pm 4) ; e=\frac{\sqrt{65}}{4}$; Latus rectum $\displaystyle =\frac{49}{2}$
    Reduce to standard form first. Divide \(\displaystyle 49y^{2}-16x^{2}=784\) throughout by \(\displaystyle 784\): \[\frac{y^{2}}{16}-\frac{x^{2}}{49}=1,\] since \(\displaystyle \dfrac{784}{49}=16\) and \(\displaystyle \dfrac{784}{16}=49\). The \(\displaystyle y^{2}\) term is positive, so the transverse axis is the \(\displaystyle y\)-axis and \[a^{2}=16,\quad b^{2}=49\ \Rightarrow\ a=4,\quad b=7.\] Using \(\displaystyle c^{2}=a^{2}+b^{2}\), \[c^{2}=16+49=65\ \Rightarrow\ c=\sqrt{65}.\]
    Foci \(\displaystyle (0,\pm\sqrt{65})\)
    Vertices \(\displaystyle (0,\pm 4)\)
    Eccentricity \(\displaystyle e=\dfrac{c}{a}=\dfrac{\sqrt{65}}{4}\)
    Latus rectum \(\displaystyle =\dfrac{2b^{2}}{a}=\dfrac{2(49)}{4}=\dfrac{49}{2}\)
    Foci \(\displaystyle (0,\pm\sqrt{65})\); vertices \(\displaystyle (0,\pm 4)\); \(\displaystyle e=\dfrac{\sqrt{65}}{4}\); latus rectum \(\displaystyle =\dfrac{49}{2}\)
  7. In each of the Exercises $\displaystyle 7$ to $\displaystyle 15$, find the equations of the hyperbola satisfying the given conditions.

    Exercise 7

    Vertices (±2,0)\displaystyle ( \pm 2,0), foci (±3,0)\displaystyle ( \pm 3,0)

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    NCERT’s answer
    $\displaystyle \frac{x^{2}}{4}-\frac{y^{2}}{5}=1$
    Standard form of a hyperbola. The vertices \(\displaystyle (\pm 2,0)\) and foci \(\displaystyle (\pm 3,0)\) lie on the \(\displaystyle x\)-axis, so the transverse axis is the \(\displaystyle x\)-axis and the equation is \[\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1.\] Comparing with vertices \(\displaystyle (\pm a,0)\) and foci \(\displaystyle (\pm c,0)\), \[a=2,\qquad c=3.\] For a hyperbola \(\displaystyle c^{2}=a^{2}+b^{2}\), so \[b^{2}=c^{2}-a^{2}=9-4=5.\]\(\displaystyle \dfrac{x^{2}}{4}-\dfrac{y^{2}}{5}=1\)

  8. Exercise 8

    Vertices (0\displaystyle 0, ± 5\displaystyle 5), foci (0\displaystyle 0, ± 8\displaystyle 8)

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    NCERT’s answer
    $\displaystyle \frac{y^{2}}{25}-\frac{x^{2}}{39}=1$
    Standard form of a hyperbola with foci on the \(\displaystyle y\)-axis.The vertices and the foci both lie on the \(\displaystyle y\)-axis, so the transverse axis is along the \(\displaystyle y\)-axis and the equation has the form \[\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1,\] with vertices \(\displaystyle (0,\pm a)\), foci \(\displaystyle (0,\pm c)\) and \(\displaystyle c^{2}=a^{2}+b^{2}\).Comparing the given data, \(\displaystyle a=5\) and \(\displaystyle c=8\). Hence \[b^{2}=c^{2}-a^{2}=8^{2}-5^{2}=64-25=39.\]\(\displaystyle \dfrac{y^{2}}{25}-\dfrac{x^{2}}{39}=1\)

  9. Exercise 9

    Vertices (0,±3)\displaystyle (0, \pm 3), foci (0,±5)\displaystyle (0, \pm 5)

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    NCERT’s answer
    $\displaystyle \frac{y^{2}}{9}-\frac{x^{2}}{16}=1$
    Standard form of a hyperbola with foci on the \(\displaystyle y\)-axis.Vertices and foci lie on the \(\displaystyle y\)-axis, so \[\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1,\qquad \text{vertices }(0,\pm a),\ \text{foci }(0,\pm c),\ c^{2}=a^{2}+b^{2}.\]Here \(\displaystyle a=3\) and \(\displaystyle c=5\), so \[b^{2}=c^{2}-a^{2}=25-9=16.\]\(\displaystyle \dfrac{y^{2}}{9}-\dfrac{x^{2}}{16}=1\)

  10. Exercise 10

    Foci (±5,0)\displaystyle ( \pm 5,0), the transverse axis is of length 8.

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    NCERT’s answer
    $\displaystyle \frac{x^{2}}{16}-\frac{y^{2}}{9}=1$
    The transverse axis has length \(\displaystyle 2a\).The foci \(\displaystyle (\pm 5,0)\) lie on the \(\displaystyle x\)-axis, so the hyperbola is \[\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1,\qquad c^{2}=a^{2}+b^{2}.\]The transverse axis (the segment joining the vertices) has length \(\displaystyle 2a\), so \[2a=8\ \Rightarrow\ a=4,\qquad c=5.\]Therefore \[b^{2}=c^{2}-a^{2}=25-16=9.\]\(\displaystyle \dfrac{x^{2}}{16}-\dfrac{y^{2}}{9}=1\)