The equation of an ellipse whose eccentricity | vedclass

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

Question

(A) The length of the major axis is the distance between the vertices $(4, 0)$ and $(10, 0)$,which is $10 - 4 = 6$.Thus,$2a = 6$,so $a = 3$.The center

Vedclass · Accepted Answer

$3x^2 + 4y^2 - 42x + 120 = 0$

Vedclass · Answer

(A) The length of the major axis is the distance between the vertices $(4, 0)$ and $(10, 0)$,which is $10 - 4 = 6$.Thus,$2a = 6$,so $a = 3$.The center of the ellipse is the midpoint of the vertices: $(\frac{4+10}{2}, 0) = (7, 0)$.Given eccentricity $e = 1/2$,we find $b^2 = a^2(1 - e^2) = 3^2(1 - (1/2)^2) = 9(3/4) = 27/4$.The equation of the ellipse is $\frac{(x - 7)^2}{a^2} + \frac{(y - 0)^2}{b^2} = 1$,which is $\frac{(x - 7)^2}{9} + \frac{y^2}{27/4} = 1$.Multiplying by $27$,we get $3(x - 7)^2 + 4y^2 = 27$.$3(x^2 - 14x + 49) + 4y^2 = 27$.$3x^2 - 42x + 147 + 4y^2 = 27$.$3x^2 + 4y^2 - 42x + 120 = 0$.

The equation of an ellipse whose eccentricity is $1/2$ and the vertices are $(4, 0)$ and $(10, 0)$ is

Similar Questions

The line $lx + my - n = 0$ will be tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,if

If the line $2x - 3y + 4 = 0$ cuts the ellipse $x = 3 \cos \theta, y = 5 \sin \theta$ at points $A$ and $B$,and $(\alpha, \beta)$ is the midpoint of $\overline{AB}$,then $3\beta - 2\alpha =$

The length of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ is $\frac{8}{3}$. If the distance from the centre of the ellipse to its focus is $\sqrt{5}$,then $\sqrt{a^2 + 6ab + b^2} =$

The coordinates of the foci of the ellipse $3x^2 + 4y^2 - 12x - 8y + 4 = 0$ are

The eccentricity of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module