Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let the eccentricity of an ellipse <img alt="\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b," src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7Bx%5E%7B2%7D%7D%7Ba%5E%7B2%7D%7D&p

Let the eccentricity of an ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b, be \frac{1}{4}. If this ellipse passes through the point \left(-4 \sqrt{\frac{2}{5}}, 3\right), then \mathrm{a}^{2}+\mathrm{b}^{2} is equal to:
 

Option: 1

29


Option: 2

31


Option: 3

32


Option: 4

34


Answers (1)

best_answer

\mathrm{e^{2}=1-\frac{b^{2}}{a^{2}}=\frac{1}{16} \Rightarrow \frac{b^{2}}{a^{2}}=\frac{15}{16}}          ............(1)

Also ellipse passes through  \mathrm{\left(-4 \sqrt{\frac{2}{5}}, 3\right)}

\mathrm{So\; \frac{32}{5 a^{2}}+\frac{9}{b^{2}}=1 \quad Put\; b^{2}=\frac{15 a^{2}}{16}}

\mathrm{\Rightarrow \frac{32}{5 a^{2}}+\frac{9}{15 a^{2}} \times 16=1 \Rightarrow \frac{80}{5 a^{2}}=1 \Rightarrow a^{2}=16 , b^{2}=15} \\

\mathrm{a^{2}+b^{2}=31 }

Hence the correct answer is option 2

Posted by

sudhir.kumar

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Latest Question