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Conic Sections
Ellipse
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Question

Let the length of the latus rectum of an ellipse with its major axis along x-axis and centre at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?

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Solution

Key Concepts and Formulas for an Ellipse

For an ellipse with its major axis along the x-axis and its center at the origin, its standard equation is given by: x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 where:

  • aa is the length of the semi-major axis (half the major axis length).
  • bb is the length of the semi-minor axis (half the minor axis length).
  • For this orientation (major axis along x-axis), we always have a>ba > b.

Other important formulas and relationships for this type of ellipse are:

  • Eccentricity (ee): A measure of how "oval" the ellipse is. It is related to aa and bb by the equation: b2=a2(1e2)b^2 = a^2(1-e^2) Alternatively, e=cae = \frac{c}{a}, where cc is the distance from the center to each focus. From b2=a2a2e2b^2 = a^2 - a^2e^2, we get a2e2=a2b2a^2e^2 = a^2 - b^2, so c2=a2b2c^2 = a^2 - b^2, which implies c=a2b2c = \sqrt{a^2 - b^2}.
  • Length of the Latus Rectum: The length of the chord passing through a focus and perpendicular to the major axis is: L=2b2aL = \frac{2b^2}{a}
  • Distance between Foci: The foci are located at (±ae,0)(\pm ae, 0). The distance between them is: Df=2aeD_f = 2ae
  • Length of the Minor Axis: The length of the minor axis is 2b2b.

Step-by-Step Derivation

Step 1: Formulate Equations from the Given Information

We are given two pieces of information about the ellipse:

  1. The length of the latus rectum is 8.

    • Using the formula for the length of the latus rectum, L=2b2aL = \frac{2b^2}{a}, we can write: 2b2a=8\frac{2b^2}{a} = 8
    • Simplifying this equation, we get our first relationship between aa and bb: b2=4a(Equation 1)b^2 = 4a \quad \text{(Equation 1)}
    • Explanation: This step translates the given verbal description into a mathematical equation using the standard formula for an ellipse.
  2. The distance between the foci is equal to the length of its minor axis.

    • Using the formula for the distance between foci, Df=2aeD_f = 2ae, and the length of the minor axis, 2b2b, we can write: 2ae=2b2ae = 2b
    • Simplifying this equation, we get our second relationship: ae=b(Equation 2)ae = b \quad \text{(Equation 2)}
    • Explanation: Again, we convert the problem statement into a mathematical equation using the relevant definitions. This equation directly relates aa, bb, and the eccentricity ee.

Step 2: Determine the Values of aa and bb

Now we have a system of equations involving aa, bb, and ee. We also know the fundamental relation b2=a2(1e2)b^2 = a^2(1-e^2). Our goal is to find aa and bb to determine the equation of the ellipse.

  1. Relate aa and bb using eccentricity:
    • From Equation 2, ae=bae = b, we can express eccentricity as e=bae = \frac{b}{a}.
    • Substitute this into the fundamental relation b2=a2(1e2)b^2 = a^2(1-e^2): b2=a2(1(ba)2)b^2 = a^2 \left( 1 - \left(\frac{b}{a}\right)^2 \right)

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