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NCERT Solutions · Class 11 Mathematics Conic Sections

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EXERCISE 10.3 1–10 (part 4 of 8)

  1. In each of the Exercises $\displaystyle 1$ to $\displaystyle 9$, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

    Exercise 1

    x236+y216=1\displaystyle \frac{x^{2}}{36}+\frac{y^{2}}{16}=1

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    NCERT’s answer
    $\displaystyle \mathrm{F}( \pm \sqrt{20}, 0) ; \mathrm{V}( \pm 6,0)$; Major axis $\displaystyle =12$; Minor axis $\displaystyle =8, e=\frac{\sqrt{20}}{6}$, Latus rectum $\displaystyle =\frac{16}{3}$
    Standard form of an ellipse. Compare \[\frac{x^{2}}{36}+\frac{y^{2}}{16}=1 \] with \(\displaystyle \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1\). Since the larger denominator, $\displaystyle 36$, sits under \(\displaystyle x^{2}\), the major axis lies along the \(\displaystyle x\)-axis, and \[a^{2}=36,\qquad b^{2}=16,\qquad\text{so } a=6,\; b=4. \]For such an ellipse \(\displaystyle c=\sqrt{a^{2}-b^{2}}\) gives the foci \(\displaystyle (\pm c,0)\), the vertices are \(\displaystyle (\pm a,0)\), the eccentricity is \(\displaystyle e=\dfrac{c}{a}\) and the latus rectum has length \(\displaystyle \dfrac{2b^{2}}{a}\).\[c=\sqrt{36-16}=\sqrt{20}=2\sqrt{5}. \]\[e=\frac{c}{a}=\frac{2\sqrt{5}}{6}=\frac{\sqrt{5}}{3},\qquad \frac{2b^{2}}{a}=\frac{2\times 16}{6}=\frac{32}{6}=\frac{16}{3}. \]Foci \(\displaystyle (\pm 2\sqrt{5},\,0)\); vertices \(\displaystyle (\pm 6,\,0)\); major axis \(\displaystyle =12\); minor axis \(\displaystyle =8\); \(\displaystyle e=\frac{\sqrt{5}}{3}\); latus rectum \(\displaystyle =\frac{16}{3}\).

  2. Exercise 2

    x24+y225=1\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{25}=1

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    NCERT’s answer
    $\displaystyle \mathrm{F}(0, \pm \sqrt{21}) ; \mathrm{V}(0, \pm 5)$; Major axis $\displaystyle =10$; Minor axis $\displaystyle =4, e=\frac{\sqrt{21}}{5}$; Latus rectum $\displaystyle =\frac{8}{5}$
    Standard form of an ellipse. Compare \[\frac{x^{2}}{4}+\frac{y^{2}}{25}=1 \] with \(\displaystyle \dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{a^{2}}=1\). Since the larger denominator, $\displaystyle 25$, sits under \(\displaystyle y^{2}\), the major axis lies along the \(\displaystyle y\)-axis, and \[a^{2}=25,\qquad b^{2}=4,\qquad\text{so } a=5,\; b=2. \]For such an ellipse \(\displaystyle c=\sqrt{a^{2}-b^{2}}\) gives the foci \(\displaystyle (0,\pm c)\), the vertices are \(\displaystyle (0,\pm a)\), the eccentricity is \(\displaystyle e=\dfrac{c}{a}\) and the latus rectum has length \(\displaystyle \dfrac{2b^{2}}{a}\).\[c=\sqrt{25-4}=\sqrt{21}=\sqrt{21}. \]\[e=\frac{c}{a}=\frac{\sqrt{21}}{5},\qquad \frac{2b^{2}}{a}=\frac{2\times 4}{5}=\frac{8}{5}. \]Foci \(\displaystyle (0,\,\pm \sqrt{21})\); vertices \(\displaystyle (0,\,\pm 5)\); major axis \(\displaystyle =10\); minor axis \(\displaystyle =4\); \(\displaystyle e=\frac{\sqrt{21}}{5}\); latus rectum \(\displaystyle =\frac{8}{5}\).

  3. Exercise 3

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

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    NCERT’s answer
    $\displaystyle \mathrm{F}( \pm \sqrt{7}, 0) ; \mathrm{V}( \pm 4,0)$; Major axis $\displaystyle =8$; Minor axis $\displaystyle =6, e=\frac{\sqrt{7}}{4}$; Latus rectum $\displaystyle =\frac{9}{2}$
    Standard form of an ellipse. Compare \[\frac{x^{2}}{16}+\frac{y^{2}}{9}=1 \] with \(\displaystyle \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1\). Since the larger denominator, $\displaystyle 16$, sits under \(\displaystyle x^{2}\), the major axis lies along the \(\displaystyle x\)-axis, and \[a^{2}=16,\qquad b^{2}=9,\qquad\text{so } a=4,\; b=3. \]For such an ellipse \(\displaystyle c=\sqrt{a^{2}-b^{2}}\) gives the foci \(\displaystyle (\pm c,0)\), the vertices are \(\displaystyle (\pm a,0)\), the eccentricity is \(\displaystyle e=\dfrac{c}{a}\) and the latus rectum has length \(\displaystyle \dfrac{2b^{2}}{a}\).\[c=\sqrt{16-9}=\sqrt{7}=\sqrt{7}. \]\[e=\frac{c}{a}=\frac{\sqrt{7}}{4},\qquad \frac{2b^{2}}{a}=\frac{2\times 9}{4}=\frac{18}{4}=\frac{9}{2}. \]Foci \(\displaystyle (\pm \sqrt{7},\,0)\); vertices \(\displaystyle (\pm 4,\,0)\); major axis \(\displaystyle =8\); minor axis \(\displaystyle =6\); \(\displaystyle e=\frac{\sqrt{7}}{4}\); latus rectum \(\displaystyle =\frac{9}{2}\).

  4. Exercise 4

    x225+y2100=1\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{100}=1

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    NCERT’s answer
    $\displaystyle \mathrm{F}(0, \pm \sqrt{75}) ; \mathrm{V}(0, \pm 10)$; Major axis $\displaystyle =20$; Minor axis $\displaystyle =10, e=\frac{\sqrt{3}}{2}$; Latus rectum = $\displaystyle 5$
    Standard form of an ellipse. Compare \[\frac{x^{2}}{25}+\frac{y^{2}}{100}=1 \] with \(\displaystyle \dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{a^{2}}=1\). Since the larger denominator, $\displaystyle 100$, sits under \(\displaystyle y^{2}\), the major axis lies along the \(\displaystyle y\)-axis, and \[a^{2}=100,\qquad b^{2}=25,\qquad\text{so } a=10,\; b=5. \]For such an ellipse \(\displaystyle c=\sqrt{a^{2}-b^{2}}\) gives the foci \(\displaystyle (0,\pm c)\), the vertices are \(\displaystyle (0,\pm a)\), the eccentricity is \(\displaystyle e=\dfrac{c}{a}\) and the latus rectum has length \(\displaystyle \dfrac{2b^{2}}{a}\).\[c=\sqrt{100-25}=\sqrt{75}=5\sqrt{3}. \]\[e=\frac{c}{a}=\frac{5\sqrt{3}}{10}=\frac{\sqrt{3}}{2},\qquad \frac{2b^{2}}{a}=\frac{2\times 25}{10}=\frac{50}{10}=5. \]Foci \(\displaystyle (0,\,\pm 5\sqrt{3})\); vertices \(\displaystyle (0,\,\pm 10)\); major axis \(\displaystyle =20\); minor axis \(\displaystyle =10\); \(\displaystyle e=\frac{\sqrt{3}}{2}\); latus rectum \(\displaystyle =5\).

  5. Exercise 5

    x249+y236=1\displaystyle \frac{x^{2}}{49}+\frac{y^{2}}{36}=1

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    NCERT’s answer
    $\displaystyle \mathrm{F}( \pm \sqrt{13}, 0) ; \mathrm{V}( \pm 7,0)$; Major axis $\displaystyle =14$; Minor axis $\displaystyle =12, e=\frac{\sqrt{13}}{7}$; Latus rectum $\displaystyle =\frac{72}{7}$
    Standard form of an ellipse. Compare \[\frac{x^{2}}{49}+\frac{y^{2}}{36}=1 \] with \(\displaystyle \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1\). Since the larger denominator, $\displaystyle 49$, sits under \(\displaystyle x^{2}\), the major axis lies along the \(\displaystyle x\)-axis, and \[a^{2}=49,\qquad b^{2}=36,\qquad\text{so } a=7,\; b=6. \]For such an ellipse \(\displaystyle c=\sqrt{a^{2}-b^{2}}\) gives the foci \(\displaystyle (\pm c,0)\), the vertices are \(\displaystyle (\pm a,0)\), the eccentricity is \(\displaystyle e=\dfrac{c}{a}\) and the latus rectum has length \(\displaystyle \dfrac{2b^{2}}{a}\).\[c=\sqrt{49-36}=\sqrt{13}=\sqrt{13}. \]\[e=\frac{c}{a}=\frac{\sqrt{13}}{7},\qquad \frac{2b^{2}}{a}=\frac{2\times 36}{7}=\frac{72}{7}. \]Foci \(\displaystyle (\pm \sqrt{13},\,0)\); vertices \(\displaystyle (\pm 7,\,0)\); major axis \(\displaystyle =14\); minor axis \(\displaystyle =12\); \(\displaystyle e=\frac{\sqrt{13}}{7}\); latus rectum \(\displaystyle =\frac{72}{7}\).

  6. Exercise 6

    x2100+y2400=1\displaystyle \frac{x^{2}}{100}+\frac{y^{2}}{400}=1

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    NCERT’s answer
    $\displaystyle \mathrm{F}(0, \pm 10 \sqrt{3}) ; \mathrm{V}(0, \pm 20)$; Major axis $\displaystyle =40$; Minor axis $\displaystyle =20, e=\frac{\sqrt{3}}{2}$; Latus rectum $\displaystyle =10$
    Standard form of an ellipse. Compare \[\frac{x^{2}}{100}+\frac{y^{2}}{400}=1 \] with \(\displaystyle \dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{a^{2}}=1\). Since the larger denominator, $\displaystyle 400$, sits under \(\displaystyle y^{2}\), the major axis lies along the \(\displaystyle y\)-axis, and \[a^{2}=400,\qquad b^{2}=100,\qquad\text{so } a=20,\; b=10. \]For such an ellipse \(\displaystyle c=\sqrt{a^{2}-b^{2}}\) gives the foci \(\displaystyle (0,\pm c)\), the vertices are \(\displaystyle (0,\pm a)\), the eccentricity is \(\displaystyle e=\dfrac{c}{a}\) and the latus rectum has length \(\displaystyle \dfrac{2b^{2}}{a}\).\[c=\sqrt{400-100}=\sqrt{300}=10\sqrt{3}. \]\[e=\frac{c}{a}=\frac{10\sqrt{3}}{20}=\frac{\sqrt{3}}{2},\qquad \frac{2b^{2}}{a}=\frac{2\times 100}{20}=\frac{200}{20}=10. \]Foci \(\displaystyle (0,\,\pm 10\sqrt{3})\); vertices \(\displaystyle (0,\,\pm 20)\); major axis \(\displaystyle =40\); minor axis \(\displaystyle =20\); \(\displaystyle e=\frac{\sqrt{3}}{2}\); latus rectum \(\displaystyle =10\).

  7. Exercise 7

    36x2+4y2=144\displaystyle 36 x^{2}+4 y^{2}=144

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    NCERT’s answer
    $\displaystyle \mathrm{F}(0, \pm 4 \sqrt{2}) ; \mathrm{V}(0, \pm 6)$; Major axis $\displaystyle =12$; Minor axis $\displaystyle =4, e=\frac{2 \sqrt{2}}{3}$; Latus rectum $\displaystyle =\frac{4}{3}$
    Standard form of an ellipse. First divide \(\displaystyle 36x^{2}+4y^{2}=144\) throughout by \(\displaystyle 144\) to reach the standard form: \[\frac{36x^{2}}{144}+\frac{4y^{2}}{144}=1\quad\Longrightarrow\quad \frac{x^{2}}{4}+\frac{y^{2}}{36}=1 . \] Compare \[\frac{x^{2}}{4}+\frac{y^{2}}{36}=1 \] with \(\displaystyle \dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{a^{2}}=1\). Since the larger denominator, $\displaystyle 36$, sits under \(\displaystyle y^{2}\), the major axis lies along the \(\displaystyle y\)-axis, and \[a^{2}=36,\qquad b^{2}=4,\qquad\text{so } a=6,\; b=2. \]For such an ellipse \(\displaystyle c=\sqrt{a^{2}-b^{2}}\) gives the foci \(\displaystyle (0,\pm c)\), the vertices are \(\displaystyle (0,\pm a)\), the eccentricity is \(\displaystyle e=\dfrac{c}{a}\) and the latus rectum has length \(\displaystyle \dfrac{2b^{2}}{a}\).\[c=\sqrt{36-4}=\sqrt{32}=4\sqrt{2}. \]\[e=\frac{c}{a}=\frac{4\sqrt{2}}{6}=\frac{2\sqrt{2}}{3},\qquad \frac{2b^{2}}{a}=\frac{2\times 4}{6}=\frac{8}{6}=\frac{4}{3}. \]Foci \(\displaystyle (0,\,\pm 4\sqrt{2})\); vertices \(\displaystyle (0,\,\pm 6)\); major axis \(\displaystyle =12\); minor axis \(\displaystyle =4\); \(\displaystyle e=\frac{2\sqrt{2}}{3}\); latus rectum \(\displaystyle =\frac{4}{3}\).

  8. Exercise 8

    16x2+y2=16\displaystyle 16 x^{2}+y^{2}=16

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    NCERT’s answer
    $\displaystyle \mathrm{F}(0, \pm \sqrt{15}) ; \mathrm{V}(0, \pm 4)$; Major axis $\displaystyle =8$; Minor axis $\displaystyle =2, e=\frac{\sqrt{15}}{4}$; Latus rectum $\displaystyle =\frac{1}{2}$
    Standard form of an ellipse. First divide \(\displaystyle 16x^{2}+y^{2}=16\) throughout by \(\displaystyle 16\) to reach the standard form: \[\frac{16x^{2}}{16}+\frac{y^{2}}{16}=1\quad\Longrightarrow\quad \frac{x^{2}}{1}+\frac{y^{2}}{16}=1 . \] Compare \[\frac{x^{2}}{1}+\frac{y^{2}}{16}=1 \] with \(\displaystyle \dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{a^{2}}=1\). Since the larger denominator, $\displaystyle 16$, sits under \(\displaystyle y^{2}\), the major axis lies along the \(\displaystyle y\)-axis, and \[a^{2}=16,\qquad b^{2}=1,\qquad\text{so } a=4,\; b=1. \]For such an ellipse \(\displaystyle c=\sqrt{a^{2}-b^{2}}\) gives the foci \(\displaystyle (0,\pm c)\), the vertices are \(\displaystyle (0,\pm a)\), the eccentricity is \(\displaystyle e=\dfrac{c}{a}\) and the latus rectum has length \(\displaystyle \dfrac{2b^{2}}{a}\).\[c=\sqrt{16-1}=\sqrt{15}=\sqrt{15}. \]\[e=\frac{c}{a}=\frac{\sqrt{15}}{4},\qquad \frac{2b^{2}}{a}=\frac{2\times 1}{4}=\frac{2}{4}=\frac{1}{2}. \]Foci \(\displaystyle (0,\,\pm \sqrt{15})\); vertices \(\displaystyle (0,\,\pm 4)\); major axis \(\displaystyle =8\); minor axis \(\displaystyle =2\); \(\displaystyle e=\frac{\sqrt{15}}{4}\); latus rectum \(\displaystyle =\frac{1}{2}\).

  9. Exercise 9

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

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    NCERT’s answer
    $\displaystyle \mathrm{F}( \pm \sqrt{5}, 0) ; \mathrm{V}( \pm 3,0)$; Major axis $\displaystyle =6$; Minor axis $\displaystyle =4, e=\frac{\sqrt{5}}{3}$; Latus rectum $\displaystyle =\frac{8}{3}$
    Standard form of an ellipse. First divide \(\displaystyle 4x^{2}+9y^{2}=36\) throughout by \(\displaystyle 36\) to reach the standard form: \[\frac{4x^{2}}{36}+\frac{9y^{2}}{36}=1\quad\Longrightarrow\quad \frac{x^{2}}{9}+\frac{y^{2}}{4}=1 . \] Compare \[\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 \] with \(\displaystyle \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1\). Since the larger denominator, $\displaystyle 9$, sits under \(\displaystyle x^{2}\), the major axis lies along the \(\displaystyle x\)-axis, and \[a^{2}=9,\qquad b^{2}=4,\qquad\text{so } a=3,\; b=2. \]For such an ellipse \(\displaystyle c=\sqrt{a^{2}-b^{2}}\) gives the foci \(\displaystyle (\pm c,0)\), the vertices are \(\displaystyle (\pm a,0)\), the eccentricity is \(\displaystyle e=\dfrac{c}{a}\) and the latus rectum has length \(\displaystyle \dfrac{2b^{2}}{a}\).\[c=\sqrt{9-4}=\sqrt{5}=\sqrt{5}. \]\[e=\frac{c}{a}=\frac{\sqrt{5}}{3},\qquad \frac{2b^{2}}{a}=\frac{2\times 4}{3}=\frac{8}{3}. \]Foci \(\displaystyle (\pm \sqrt{5},\,0)\); vertices \(\displaystyle (\pm 3,\,0)\); major axis \(\displaystyle =6\); minor axis \(\displaystyle =4\); \(\displaystyle e=\frac{\sqrt{5}}{3}\); latus rectum \(\displaystyle =\frac{8}{3}\).
  10. In each of the following Exercises $\displaystyle 10$ to $\displaystyle 20$, find the equation for the ellipse that satisfies the given conditions:

    Exercise 10

    Vertices (±5,0)\displaystyle ( \pm 5,0), foci (±4,0)\displaystyle ( \pm 4,0)

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    NCERT’s answer
    $\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
    Standard form of an ellipse. The vertices and foci both lie on the \(\displaystyle x\)-axis, so the major axis is along the \(\displaystyle x\)-axis and the equation has the form \[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\qquad a>b.\] Here the vertices are \(\displaystyle (\pm a,0)\) and the foci are \(\displaystyle (\pm c,0)\), so \[a=5,\qquad c=4.\] For an ellipse \(\displaystyle c^{2}=a^{2}-b^{2}\), hence \[b^{2}=a^{2}-c^{2}=25-16=9.\] So \(\displaystyle a^{2}=25\) and \(\displaystyle b^{2}=9\).\(\displaystyle \dfrac{x^{2}}{25}+\dfrac{y^{2}}{9}=1\)