JEE Main 2020 · 6 September · evening · Single correct · +4 / −1

If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity e of the ellipse satisfies :

  1. (a) e⁴ + 2e² – 1 = 0
  2. (b) e⁴ + e² – 1 = 0
  3. (c) e² + 2e – 1 = 0
  4. (d) e² + e – 1 = 0
Answer and solution

Answer (b) e⁴ + e² – 1 = 0

Solution

Maths sir · 4 steps

Find where the normal at (ae, b²/a) cuts the y-axis, set that equal to -b, then remove b.

  1. Slope of normal. Take x²/a²+y²/b²=1 and the latus rectum end P(ae, b²/a). The tangent slope there is -(b² x)/(a² y) = -(b² · ae)/(a² · b²/a) = -e, so the normal has slope 1/e.
  2. Meet the y-axis. Put x = 0 in the normal through P: y = b²/a - 1/e(ae) = (b² - a²)/a = -a²e²/a = -ae². This is negative, so the normal passes through the lower extremity (0, -b).

    -ae² = -b ⇒ b = ae²

  3. Remove b. Square both sides and use b² = a²(1-e²): a²e⁴ = a²(1-e²), so e⁴ + e² - 1 = 0.
  4. So. The eccentricity satisfies e⁴ + e² - 1 = 0, which is option (b).

The idea to keep. For a normal through a given point, find where it meets the line you need, write the condition, then replace b² by a²(1-e²) to get an equation in e alone.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Ellipse

Asked about 100 times in JEE Main since 2020. This question is from 2020.