JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Ellipse

Consider two straight lines, each of which is tangent to both the circle

x^{2} + y^{2} = \frac{1}{2}

and the parabola

y^{2} = 4 x

. Let these lines intersect at the point

Q

. Consider the ellipse whose center is at the origin

O (0, 0)

and whose semi-major axis is

O Q

. If the length of the minor axis of this ellipse is

2

, then which of the following statement(s) is (are) TRUE?

(A)For the ellipse, the eccentricity is $\frac{1}{2}$ and the length of the latus rectum is 1✓ Correct
(B)For the ellipse, the eccentricity is $\frac{1}{2}$ and the length of the latus rectum is $\frac{1}{2}$
(C)The area of the region bounded by the ellipse between the lines $x = \frac{1}{2}$ and $x = 1$ is $\frac{1}{4 2} (π - 2)$ ✓ Correct
(D)The area of the region bounded by the ellipse between the lines $x = \frac{1}{2}$ and $x = 1$ is $\frac{1}{16} (π - 2)$

Answer: (A), (C)

Solution

Common tangent to circle and parabola:

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

and

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

Equating:

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

The tangent lines are

y = x + 1

and

y = - x - 1

, intersecting at

Q (- 1, 0)

.

So

a = ∣ O Q ∣ = 1

. Given

2 b = 2

, so

b = \frac{1}{2}

.

Eccentricity:

e = 1 - \frac{b ^{2}}{a ^{2}} = 1 - \frac{1/2}{1} = \frac{1}{2}

✓

Latus rectum:

\frac{2 b ^{2}}{a} = \frac{2 \cdot 1/2}{1} = 1

✓

Ellipse equation:

\frac{x ^{2}}{1} + \frac{y ^{2}}{1/2} = 1

Area =

2 \int_{1/ 2}^{1} \frac{1}{2} 1 - x^{2} d x = 2 [\frac{x}{2} 1 - x^{2} + \frac{1}{2} sin^{- 1} x]_{1/ 2}^{1} = \frac{π - 2}{4 2}

✓

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · One or more correct options