Find the equation for the ellipse that satisf | vedclass

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

Question

(A) The vertices are $(\pm 5, 0)$,which lie on the $x$-axis. Thus,the equation is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Here,$a = 5$ and

Vedclass · Accepted Answer

$\frac{x^2}{25} + \frac{y^2}{9} = 1$

Vedclass · Answer

(A) The vertices are $(\pm 5, 0)$,which lie on the $x$-axis. Thus,the equation is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Here,$a = 5$ and the distance of the foci from the center is $c = 4$.Using the relation $a^2 = b^2 + c^2$,we have $5^2 = b^2 + 4^2$.$25 = b^2 + 16 \implies b^2 = 9$.Substituting $a^2 = 25$ and $b^2 = 9$ into the standard equation,we get $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 5, 0)$,foci $(\pm 4, 0)$.

Similar Questions

The equation of the tangent to the ellipse $x^2 + 16y^2 = 16$ making an angle of $60^\circ$ with the $x$-axis is

Find the coordinates of the foci,the vertices,the lengths of the major and minor axes,and the eccentricity of the ellipse $9x^{2} + 4y^{2} = 36$.

At which points on the ellipse $4x^2 + 9y^2 = 1$ is the tangent line parallel to the line $8x = 9y$?

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

The locus of the feet of the perpendiculars from either focus of the ellipse $(x - y + 1)^2 + (2x + 2y - 6)^2 = 20$ onto any tangent is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module