Q.Find the equation of the ellipse referred to its major and minor axes as the coordinate axes X, Y-respectively with latus rectum of length 4, and distance between foci $4\sqrt{2}$.

Showing the 12 most recent of 18 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.Eccentricity of the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ is

   (a) $\dfrac{\sqrt{7}}{2}$

   (b) $\dfrac{5}{3}$

   (c) $\dfrac{\sqrt{7}}{3}$

   (d) $\dfrac{\sqrt{7}}{4}$

   ›Reveal solutionSolution

   Identify $a^2$ (larger denominator, major axis) and $b^2$, then apply $e=\sqrt{1-b^2/a^2}$.

   For $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$: since $16>9$, the major axis is along the $x$-axis, with $a^2=16$ and $b^2=9$.

   …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Assertion (A): The eccentricity of ellipse x²/16 + y²/9 = 1 is 5/4. Reason (R): The eccentricity of ellipse x²/a² + y²/b² = 1 is given by b² = a²(1 - e²).

   (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

   (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

   (c) Assertion (A) is true and Reason (R) is false.

   (d) Assertion (A) is false and Reason (R) is true.

   ›Reveal solutionSolution

   Compute the real eccentricity using b² = a²(1−e²); it comes out to √7/4, not the impossible value 5/4 claimed in the Assertion — so Assertion is false, Reason (the formula) is true.

   Checking the Assertion (A): For the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$, we have $a^2=16$, $b^2=9$ (and $a^2>b^2$, so the major axis is along the x-axis).

   Using the standard relation $b^2=a^2(1-e^2)$:

   $9 = 16(1-e^2) \implies 1-e^2 = \frac{9}{16} \implies e^2 = \frac{7}{16} \implies e = \frac{\sqrt7}{4} \approx 0.66$

   This is nowhere near $\dfrac54=1.25$. In fact, for any real ellipse the eccentricity must satisfy $0<e<1$, so $e=\dfrac54$ is impossible on its face. Assertion (A) is false.

   …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2026Set ANNUAL1 markMCQ
   Q.Length of Latus rectum of an ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ is

   (a) $\dfrac{2a^2}{b}$

   (b) $2a$

   (c) $2b$

   (d) $\dfrac{2b^2}{a}$

   ›Reveal solutionSolution

   For the standard ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with the major axis along the x-axis ($a>b$), the latus rectum through each focus has length $\dfrac{2b^2}{a}$.

   The latus rectum of an ellipse is the chord through a focus, perpendicular to the major axis. For the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a>b$, substituting $x=\pm ae$ (the focus) into the ellipse equation and solving for $y$ gives …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2025Set ANNUAL1 markMCQ
   Q.The length of the minor axis of the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{25} = 1$ is

   (a) 4

   (b) 8

   (c) 5

   (d) 10

   ›Reveal solutionSolution

   Minor axis of $\dfrac{x^2}{16}+\dfrac{y^2}{25}=1$ is $8$.

   …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2025Set ANNUAL1 markMCQ
   Q.The vertices of an ellipse are $(\pm 5, 0)$ and its foci are $(\pm 4, 0)$. The equation of the ellipse is

   (a) $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$

   (b) $\dfrac{x^2}{9} + \dfrac{y^2}{25} = 1$

   (c) $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$

   (d) $\dfrac{x^2}{16} + \dfrac{y^2}{25} = 1$

   ›Reveal solutionSolution

   With vertices at $(\pm5,0)$ giving $a=5$ and foci at $(\pm4,0)$ giving $c=4$, the relation $b^2=a^2-c^2$ gives $b^2=9$, so the ellipse is $\frac{x^2}{25}+\frac{y^2}{9}=1$.

   For an ellipse with major axis along the x-axis, standard form is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a>b$.

   Vertices $(\pm5,0) \implies a=5 \implies a^2=25$

   Foci $(\pm4,0) \implies c=4$

   …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2025Set sz1 mark
   Q.The coordinates of the foci of the ellipse $\dfrac{x^2}{100} + \dfrac{y^2}{9} = 1$ are ............... .

   ›Reveal solutionSolution

   With $a^2=100$ (under $x^2$) and $b^2=9$ (under $y^2$), the ellipse's major axis is horizontal and its foci lie at $(\pm\sqrt{91}, 0)$.

   The standard form $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a^2 > b^2$ has its major axis along the x-axis, foci at $(\pm c, 0)$ where $c^2 = a^2-b^2$ …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2024Set ANNUAL1 markMCQ
   Q.The foci of the ellipse $\dfrac{x^2}{169}+\dfrac{y^2}{25}=1$ are

   (a) $(\pm 12, 0)$

   (b) $(0, \pm\sqrt{20})$

   (c) $(\pm\sqrt{20}, 0)$

   (d) None of these

   ›Reveal solutionSolution

   Identify $a^2$ and $b^2$ from the ellipse equation (larger denominator under $x^2$ means the major axis is horizontal), then use $c^2=a^2-b^2$ to find the foci.

   The ellipse is $\dfrac{x^2}{169}+\dfrac{y^2}{25}=1$. Since $169 > 25$, the major axis lies along the $x$-axis, with $a^2 = 169$ and $b^2 = 25$.

   For an ellipse, the distance from centre to each focus is: …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2024Set ANNUAL1 mark
   Q.Write equation of directrix of an ellipse x^2/a^2 + y^2/b^2 = 1 (a > b).

   ›Reveal solutionSolution

   For $a>b$ the major axis is along the x-axis, so the directrices are vertical lines $x=\pm a/e$.

   For the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a>b$, the major axis lies along the $x$-axis and the two foci are at $(\pm ae, 0)$, where the eccentricity is

   $e = \sqrt{1-\dfrac{b^2}{a^2}}$.

   …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2024Set ANNUAL1 markMCQ
   Q.The foci of Ellipse $\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$ is:

   (a) $(0, 0)$

   (b) $(\pm ae, 0)$

   (c) $(0, \pm ae)$

   (d) None of these.

   ›Reveal solutionSolution

   For $\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$ with $a>b$, the foci are $(0,\pm ae)$.

   In this equation, the larger denominator $a^2$ sits under $y^2$, so the major axis of the ellipse is along the $y$-axis (this is a vertically oriented ellipse). For such an ellipse, the foci lie on the major axis at a distance $ae$ from the centre, where $e$ is the eccentri …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2024Set sz1 markMCQ
    Q.$e$ (eccentricity) of ellipse is:

    (a) $e < 1$

    (b) $e = 1$

    (c) $e > 1$

    (d) $e = 2$

    ›Reveal solutionSolution

    An ellipse is defined by eccentricity $e$ satisfying $0 < e < 1$.

    Eccentricity classifies conic sections:

    • Circle: $e = 0$
    • Ellipse: $0 < e < 1$
    • Parabola: $e = 1$
    • Hyperbola: $e > 1$ …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2024Set hz1 mark
    Q.Eccentricity of the ellipse $\dfrac{x^2}{36} + \dfrac{y^2}{16} = 1$ is .......

    ›Reveal solutionSolution

    Comparing to the standard ellipse form gives $a^2=36$, $b^2=16$; the eccentricity formula $e=\sqrt{1-b^2/a^2}$ then gives $e=\sqrt5/3$.

    The given ellipse is $\dfrac{x^2}{36} + \dfrac{y^2}{16} = 1$.

    Since the denominator under $x^2$ is larger, the major axis is along the $x$-axis, so $a^2 = 36$ and $b^2 = 16$, i.e. $a=6$, $b=4$.

    …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2023Set ANNUAL1 markMCQ
    Q.The eccentricity of the ellipse $\dfrac{x^2}{36} + \dfrac{y^2}{16} = 1$ is

    (a) $\dfrac{6}{\sqrt{20}}$

    (b) $\dfrac{\sqrt{6}}{20}$

    (c) $\dfrac{\sqrt{20}}{6}$

    (d) $\dfrac{20}{\sqrt{6}}$

    ›Reveal solutionSolution

    Since $a^2=36>b^2=16$, the major axis is along the $x$-axis, and $e=\sqrt{1-b^2/a^2}=\sqrt{20}/6$.

    For the ellipse $\dfrac{x^2}{36}+\dfrac{y^2}{16}=1$, since $36>16$, the major axis is along the $x$-axis, so $a^2=36$, $b^2=16$.

    Eccentricity formula: …

    🔒 Start your 14-day free trial to see the full solution →