Let the common tangents to the curves 4(x² + y²) = 9 and y² = 4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and l respectively denote the eccentricity and the length of the latus rectum of this ellipse, then l/e² is equal to ______________.

Solution

Maths sir · 5 steps

Write the parabola's tangent in slope form, then make it touch the circle. That gives Q.

1. Tangent to parabola. For y² = 4x we have a = 1, so every tangent has the form y = mx + 1/m. The circle is x² + y² = 9/4, with centre O and radius 3/2.
2. Touch the circle. The line is also tangent to the circle when its distance from O equals the radius: (|1/m|)/√(1+m²) = 3/2. Squaring and clearing fractions gives a quadratic in m².

   9m⁴ + 9m² - 4 = 0 ⇒ (3m² - 1)(3m² + 4) = 0 ⇒ m² = 1/3

3. Find Q. With m = ±1/√3, the common tangents are y = x/√3 + √3 and y = -x/√3 - √3. They are mirror images in the x-axis, so they meet on it, where x = -3. So Q = (-3, 0) and OQ = 3.
4. Build the ellipse. Here b = 3 and a = 6, so e² = 1 - b²/a² = 1 - 9/36 = 3/4. The latus rectum is l = 2b²/a = 18/6 = 3.
5. So. l/e² = 3/(3/4) = 4.

The common slip: 2. Using the semi-latus rectum b²/a = 3/2 instead of the full 2b²/a = 3 gives 2, not 4.

The idea to keep. For a common tangent to a parabola and a circle, take the parabola's slope form y = mx + a/m and set its distance from the circle's centre equal to the radius.

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