Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The eccentricity of the ellipse which meets the straight line <img alt="\mathrm{\frac{x}{7}+\frac{y}{2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%7D%7B7%7D&am

The eccentricity of the ellipse which meets the straight line \mathrm{\frac{x}{7}+\frac{y}{2}=1} on the axis of x and the straight line \mathrm{e^{\frac{x}{3}-\frac{y}{5}=1}} on the axis of y and whose axes lie along the axes of coordinates is
 

Option: 1

\frac{3 \sqrt{2}}{7}


Option: 2

\frac{2 \sqrt{6} }{7}


Option: 3

\frac{ \sqrt{3} }{7}


Option: 4

none of these


Answers (1)

best_answer

Let the equation of the ellipse be \mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}
Which passes through the points (7,0) and (0,-5). So \mathrm{ a^2=49} and \mathrm{b^2=25}.
\mathrm{a a^2>b^2}
\mathrm{\therefore b^2=a^2\left(1-e^2\right) \Rightarrow 25=49\left(1-e^2\right) \Rightarrow e=\frac{2 \sqrt{6} }{7}. }Hence (b) is the correct answer.

Posted by

vinayak

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today
anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question