Consider two straight lines, each of which is tangent to both the circle x ^ 2 + y ^ 2 = (1)/(2) and the parabola y ^ 2 = 4x . Let these lines intersect at the point Q . Consider the ellipse whose center is at the

Correct answer: A. For the ellipse, the eccentricity is 1 2 and the length of the latus rectum is 1 Let equation of common tangent is ( y=m x+ 1 ~m ( | 0+0+ 1 ~m 1+ ^ 2 |= 1 2 ^ 4 + ^ 2 - 2=0 m=± 1) Equation of common tangents are ( y=x+ 1) and ( y=-x+ 1) point Q is (-1,0) ( ) Equation of ellipse is ( x^ 2 1 + y^ 2 1 / 2 = 1) (A) ( e= 1-(1)/(2) = 1 2 and ( LR= 2 ~b^ 2 a = 1) (C) Area (2 · ∫_ 1 / 2 ^ 1 1 2 · 1- ^ 2 dx= 2 [ x 2 1- ^ 2 +(1)/(2) ^ -1 x]_ 1 / 2 ^ 1 (= 2 [(π)/(4)-((1)/(4)+(π)/(8))]= 2 ((π)/(8)-(1)/(4)]= π-2 4 2 Correct answers are (A) and (C).

JEE Advanced 2018Paper 2

Exam: JEE Advanced
Year: 2018
Paper: Paper 2
Subject: Mathematics
Chapter: Ellipse

JEE Advanced 2018 · Paper 2 · Mathematics · Ellipse

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

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

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

and the parabola

{y}^{2} = 4x

. 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

OQ

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

\sqrt{2}

, then the which of the following statement(s) is (are) TRUE? Consider two straight lines, each of which is tangent to both the circle x ^ 2 + y ^ 2 = (1)/(2) and the parabola y ^ 2

Consider two straight lines, each of which is tangent to both the circle x ^ 2 + y ^ 2 = (1)/(2) and the parabola y ^ 2

Options

AFor the ellipse, the eccentricity is $\dfrac{1}{\sqrt{2}}$ and the length of the latus rectum is 1
BFor the ellipse, the eccentricity is $\dfrac{1}{2}$ and the length of the latus rectum is $\dfrac{1}{2}$
CThe area of the region bounded by the ellipse between the lines $x = \dfrac{1}{\sqrt{2}} and$ x = 1 $is$ \dfrac{1}{4\sqrt{2}}\left(\pi - 2\right)
DThe area of the region bounded by the ellipse between the lines $x = \dfrac{1}{\sqrt{2}} and$ x = 1 $is$ \dfrac{1}{16}\left(\pi - 2\right)

Correct answer

A. For the ellipse, the eccentricity is $\dfrac{1}{\sqrt{2}}$ and the length of the latus rectum is 1

Step-by-step solution

Let equation of common tangent is (

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

(\therefore

\left|\frac{0+0+\frac{1}{\mathrm{~m}}}{\sqrt{1+\mathrmm^{2}}}\right|=\frac{1}{\sqrt{2}} \Rightarrow \mathrmm^{4}+\mathrmm^{2}-

2=0 \Rightarrow

\mathrm{m}=\pm

1) Equation of common tangents are (

\mathrm{y}=\mathrm{x}+

1) and (

\mathrm{y}=-\mathrm{x}+

1) point

Q

(-1,0)

(\therefore) Equation

of ellipse is (\frac{x^{2}}{1}+\frac{y^{2}}{1 / 2}=

1) (A) (

\mathrm{e}=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}

and (

\mathrm{LR}=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=

1) (C) Area (2

\cdot \int_{1 / \sqrt{2}}^{1} \frac{1}{\sqrt{2}} \cdot \sqrt{1-\mathrmx^{2}} \mathrm{dx}=\sqrt{2}\left[\frac{\mathrm{x}}{2} \sqrt{1-\mathrmx^{2}}+\frac{1}{2} \sin ^{-1} \mathrm{x}\right]_{1 / \sqrt{2}}^{1}

\sqrt{2}\left[\frac{\pi}{4}-\left(\frac{1}{4}+\frac{\pi}{8}\right)\right]=\sqrt{2}\left(\frac{\pi}{8}-\frac{1}{4}\right]=\frac{\pi-2}{4 \sqrt{2}}

Correct answers are (A) and (C). Consider two straight lines, each of which is tangent to both the circle x ^ 2 + y ^ 2 = (1)/(2) and the parabola y ^ 2

Practice Ellipse on Quantrex Academy →

More from Ellipse