The distance between the foci of the ellipse | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given equation of the ellipse is $3x^2 + 4y^2 = 48$.Dividing both sides by $48$,we get $\frac{x^2}{16} + \frac{y^2}{12} = 1$.Comparing this wi

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The given equation of the ellipse is $3x^2 + 4y^2 = 48$.Dividing both sides by $48$,we get $\frac{x^2}{16} + \frac{y^2}{12} = 1$.Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 16$ and $b^2 = 12$.Thus,$a = 4$ and $b = \sqrt{12} = 2\sqrt{3}$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{12}{16}} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.The distance between the foci is $2ae = 2 	imes 4 	imes \frac{1}{2} = 4$.

The distance between the foci of the ellipse $3x^2 + 4y^2 = 48$ is

Similar Questions

From point $P(8, 27)$,tangents $PQ$ and $PR$ are drawn to the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$. The angle subtended by $QR$ at the origin is:

The equation of a chord $AB$ of an ellipse $2x^2 + y^2 = 1$ is $x - y + 1 = 0$. If $O$ is the origin,then $\angle AOB =$

If $4x+y+p=0$ $(p>0)$ is a tangent to the ellipse $x^2+3y^2=3$ and $16x+qy+14=0$ $(q>0)$ is a normal to the ellipse $x^2+8y^2=33$,then $p+q=$

Let $S$ and $S'$ be the foci of an ellipse and $B$ be any one of the extremities of its minor axis. If $\Delta S'BS$ is a right-angled triangle with the right angle at $B$ and $\text{Area}(\Delta S'BS) = 8 \text{ sq. units}$,then the length of the latus rectum of the ellipse is

The number of real tangents that can be drawn to the ellipse $3x^2 + 5y^2 = 32$ passing through $(3, 5)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module