The equations of the directrices of the ellip | vedclass

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

Question

(D) The given equation of the ellipse is $16x^2 + 25y^2 = 400$.Dividing by $400$,we get $\frac{x^2}{25} + \frac{y^2}{16} = 1$.Here,$a^2 = 25$ and $b^2

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The given equation of the ellipse is $16x^2 + 25y^2 = 400$.Dividing by $400$,we get $\frac{x^2}{25} + \frac{y^2}{16} = 1$.Here,$a^2 = 25$ and $b^2 = 16$,so $a = 5$ and $b = 4$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.The equations of the directrices for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are $x = \pm \frac{a}{e}$.Substituting the values,$x = \pm \frac{5}{3/5} = \pm \frac{25}{3}$,which implies $3x = \pm 25$.Since $3x = \pm 25$ is not among the options $A, B, C$,the correct choice is $D$.

The equations of the directrices of the ellipse $16x^2 + 25y^2 = 400$ are

Similar Questions

The eccentricity of the curve represented by the equation ${x^2} + 2{y^2} - 2x + 3y + 2 = 0$ is

What is the equation of the ellipse with foci $(\pm 2, 0)$ and eccentricity $e = \frac{1}{2}$?

The angle between the pair of tangents drawn from the point $(1, 2)$ to the ellipse $3x^2 + 2y^2 = 5$ is:

The product of the perpendiculars drawn from the points $(\pm \sqrt{a^2 - b^2}, 0)$ to the line $\frac{x}{a}\cos \theta + \frac{y}{b}\sin \theta = 1$ is:

If $6x - 5y - 20 = 0$ is a normal to the ellipse $x^2 + 3y^2 = k$,then $k =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module