Let the equations of two ellipses be E_1: (x^2)/(3)+(y^2)/(2)= 1 and E_2: (x^2)/(16)+(y^2)/(b^2)= 1 , If the product of their eccentricities is (1)/(2) , then the length of the minor axis of ellipse E_2 is :

Correct answer: C. 4 Given equations of ellipses aligned E_ 1 : (x^2)/(3)+(y^2)/(2)=1 & e_1= 1-(2)/(3) = 1 3 & and E_ 2 : (x^2)/(16)+(y^2)/(b^2)=1 & e_2= (1-b^2)/(16) = (16-b^2)/(4) Also, given e_1 × e_2=(1)/(2) & 1 3 × (16-b^2)/(4) =(1)/(2) 16-b²=12 & b²=4 & Length of minor axis of & E_ 2 =2 b=2 ×2=4 aligned

JEE Main 201322 AprilOnline

Exam: JEE Main
Year: 2013
Date: 22 April
Mode: Online
Subject: Mathematics
Chapter: Ellipse

JEE Main 2013 · 22 April · Online · Mathematics · Ellipse

Let the equations of two ellipses be E_1: (x^2)/(3)+(y^2)/(2)= 1...

Let the equations of two ellipses be E_1:

\frac{x^2}{3}+\frac{y^2}{2}=

\text{ and }

E_2:

\frac{x^2}{16}+\frac{y^2}{b^2}=

\text{, }

If the product of their eccentricities is

\frac{1}{2}

, then the length of the minor axis of ellipse

E_2

is :

Options

Correct answer

C. 4

Step-by-step solution

Given equations of ellipses

\begin{aligned} \mathrm{E}_{1}: \frac{x^2}{3}+\frac{y^2}{2}=1 \\ & \Rightarrow e_1=\sqrt{1-\frac{2}{3}}=\frac{1}{\sqrt{3}}\\ & \text{ and } \mathrm{E}_{2}: \frac{x^2}{16}+\frac{y^2}{b^2}=1 \\ & \Rightarrow e_2=\sqrt{\frac{1-b^2}{16}}=\sqrt{\frac{16-b^2}{4}}\\ \text{ Also, given } e_1 \times e_2=\frac{1}{2}\\ & \Rightarrow \frac{1}{\sqrt{3}} \times \sqrt{\frac{16-b^2}{4}}=\frac{1}{2} \Rightarrow16-b²=12 \\ & \Rightarrow b²=4 \\ & \therefore \text{ Length of minor axis of }\\ & \mathrm{E}_{2}=2 b=2 \times2=4 \end{aligned}

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