JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Ellipse

Let the common tangents to the curves

4 (x^{2} + y^{2}) = 9

and

y^{2} = 4 x

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

\frac{l}{e ^{2}}

is equal to ______.

Answer: 4

Solution

The circle is

x^{2} + y^{2} = \frac{9}{4}

and the parabola is

y^{2} = 4 x

.

For the circle, tangent in slope form:

y = m x \pm \frac{3}{2} 1 + m^{2}

For the parabola:

y = m x + \frac{1}{m}

Equating:

\pm \frac{3}{2} 1 + m^{2} = \frac{1}{m}

9 m^{2} (1 + m^{2}) = 4

9 m^{4} + 9 m^{2} - 4 = 0

(3 m^{2} + 4) (3 m^{2} - 1) = 0

m^{2} = \frac{1}{3}

, so

m = \pm \frac{1}{3}

Common tangent:

y = \frac{x}{3} + 3

y = 0

x = - 3

, so

Q = (- 3, 0)

and

O Q = 3

.

For the ellipse:

b = 3

a = 6

e^{2} = 1 - \frac{b ^{2}}{a ^{2}} = 1 - \frac{9}{36} = \frac{3}{4}

l = \frac{2 b ^{2}}{a} = \frac{2 \times 9}{6} = 3

\frac{l}{e ^{2}} = \frac{3}{3/4} = 4

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Ellipse

Solution

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