Equation of the ellipse with eccentricity $\frac{1}{2}$ and foci at $( \pm 1,0)$ is
40 views
0 Votes
0 Votes

Equation of the ellipse with eccentricity $\frac{1}{2}$ and foci at $( \pm 1,0)$ is

(A) $\frac{x^2}{3}+\frac{y^2}{4}=1$

(B) $\frac{x^2}{4}+\frac{y^2}{3}=1$

(C) $\frac{x^2}{4}+\frac{y^2}{3}=\frac{4}{3}$

(D) None of these

1 Answer

0 Votes
0 Votes
 
Best answer

SOLUTION — Here,

$e=\frac{1}{2} \text { and } a e= \pm 1 \Rightarrow a= \pm 2$

$\begin{array}{ll}\therefore b^2=a^2\left(1-e^2\right)=4\left(1-\frac{1}{4}\right)=3 \\\therefore \text { Equation is } \frac{x^2}{4}+\frac{y^2}{3}=1\end{array}$

So, The correct option will be (B).

RELATED DOUBTS

1 Answer
0 Votes
0 Votes
51 Views
1 Answer
0 Votes
0 Votes
122 Views
1 Answer
0 Votes
0 Votes
101 Views
1 Answer
0 Votes
0 Votes
94 Views
1 Answer
0 Votes
0 Votes
96 Views
1 Answer
0 Votes
0 Votes
128 Views
1 Answer
0 Votes
0 Votes
84 Views
1 Answer
0 Votes
0 Votes
87 Views
1 Answer
0 Votes
0 Votes
100 Views
1 Answer
1 Vote
1 Vote
91 Views
1 Answer
0 Votes
0 Votes
108 Views
Peddia is an Online Question and Answer Website, That Helps You To Prepare India's All States Boards & Competitive Exams Like IIT-JEE, NEET, AIIMS, AIPMT, SSC, BANKING, BSEB, UP Board, RBSE, HPBOSE, MPBSE, CBSE & Other General Exams.
If You Have Any Query/Suggestion Regarding This Website or Post, Please Contact Us On : [email protected]

CATEGORIES