The equation of the ellipse whose foci are (± ,2,0) and eccentricity (1/2) is:

Question

The equation of the ellipse whose foci are (± ,2,0) and eccentricity (1/2) is: (A)  (x2/12)+(y2/16)=1  (B)  (x2/16)+(y2/12)=1  (C)  (x2/16)+(y2/8)=1  (D) none of these

Tardigrade Admin · Accepted Answer

Let the ellipse be (x2/a2)+(y2/b2)=1 It is given that e=(1/2) and ae=2 Therefore, a=4 Now, b2=a2(1-e2) ⇒ b2=12 Thus the required ellipse is (x2/16)+(y2/12)=1.

Q. The equation of the ellipse whose foci are $(\pm, 2, 0)$ and eccentricity $\frac{1}{2}$ is:

$\frac{x ^{2}}{12} + \frac{y ^{2}}{16} = 1$

$\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1$

$\frac{x ^{2}}{16} + \frac{y ^{2}}{8} = 1$

none of these

Let the ellipse be $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ It is given that $e = \frac{1}{2}$ and $a e = 2$ Therefore, $a = 4$ Now, $b^{2} = a^{2} (1 - e^{2})$ $\Rightarrow$ $b^{2} = 12$ Thus the required ellipse is $\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1.$

Q. The equation of the ellipse whose foci are (±,2,0) and eccentricity 21​ is:

12x2​+16y2​=1

16x2​+12y2​=1

16x2​+8y2​=1