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NCERT Exemplar · Q49

Q.The equation of the ellipse having foci (0,1)(0, 1), (0,−1)(0, -1) and minor axis of length 1 is ________.

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We determine the center and orientation from the foci, use the minor axis length to find bb, and then use the relation b2=a2−c2b^2 = a^2 - c^2 to find aa. The equation of the ellipse is 20x2+4y2=5\boxed{20x^2 + 4y^2 = 5}.

The equation of an ellipse is fundamentally derived from its definition as a locus of points. An ellipse is the set of all points in a plane such that the sum of the distances from two fixed points (called foci) is constant. This constant sum is equal to the length of the major axis, denoted as 2a2a.

The key to finding the equation of an ellipse is to identify its center, the lengths of its semi-major axis (aa) and semi-minor axis (bb), and its orientation (whether the major axis is horizontal or vertical). These parameters are interconnected by the relationship b2=a2−c2b^2 = a^2 - c^2, where cc is the distance from the center to each focus.

Let's break down the problem:

  1. Identify the Foci and Center:

    The given foci are F1(0,1)F_1(0, 1) and F2(0,−1)F_2(0, -1).

    The center of the ellipse is the midpoint of the segment connecting the foci.

    Center C=(0+02,1+(−1)2)=(0,0)C = \left(\frac{0+0}{2}, \frac{1+(-1)}{2}\right) = (0, 0).

    The distance between the foci is 2c2c.

    2c=(0−0)2+(1−(−1))2=02+22=4=22c = \sqrt{(0-0)^2 + (1-(-1))^2} = \sqrt{0^2 + 2^2} = \sqrt{4} = 2.

    Therefore, c=1c = 1.

    Since the foci lie on the y-axis, the major axis of the ellipse is vertical. This means the standard form of the ellipse equation will be x2b2+y2a2=1\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1.

  2. Determine the Semi-minor Axis Length:

    The problem states that the minor axis has a length of 1.

    The length of the minor axis is 2b2b.

    So, 2b=12b = 1, which implies b=12b = \frac{1}{2}.

    Then, b2=(12)2=14b^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}.

  3. Calculate the Semi-major Axis Length:

    We use the fundamental relationship between aa, bb, and cc for an ellipse:

    b2=a2−c2b^2 = a^2 - c^2

    Substitute the values we found for b2b^2 and cc:

    14=a2−12\frac{1}{4} = a^2 - 1^2

    14=a2−1\frac{1}{4} = a^2 - 1

    a2=1+14a^2 = 1 + \frac{1}{4}

    a2=54a^2 = \frac{5}{4}. …

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