An ellipse having the coordinate axes as its | vedclass

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

Question

(D) Since the major axis is along the $Y$-axis,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $b > a$. Given eccentricit

Vedclass · Accepted Answer

$5x^2 + 3y^2 - 48 = 0$

Vedclass · Answer

(D) Since the major axis is along the $Y$-axis,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $b > a$. Given eccentricity $e = \sqrt{\frac{2}{5}}$,we have $e^2 = 1 - \frac{a^2}{b^2} = \frac{2}{5}$. $\frac{a^2}{b^2} = 1 - \frac{2}{5} = \frac{3}{5} \implies b^2 = \frac{5a^2}{3}$. The ellipse passes through $(-3, 1)$,so $\frac{(-3)^2}{a^2} + \frac{1^2}{b^2} = 1$. $\frac{9}{a^2} + \frac{1}{b^2} = 1$. Substituting $b^2 = \frac{5a^2}{3}$,we get $\frac{9}{a^2} + \frac{3}{5a^2} = 1$. $\frac{45 + 3}{5a^2} = 1 \implies 5a^2 = 48 \implies a^2 = \frac{48}{5}$. Then $b^2 = \frac{5}{3} 	imes \frac{48}{5} = 16$. The equation is $\frac{x^2}{48/5} + \frac{y^2}{16} = 1$. $\frac{5x^2}{48} + \frac{y^2}{16} = 1 \implies 5x^2 + 3y^2 = 48$.

An ellipse having the coordinate axes as its axes and its major axis along the $Y$-axis,passes through the point $(-3, 1)$ and has eccentricity $e = \sqrt{\frac{2}{5}}$. Then its equation is:

Similar Questions

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 $p+q=$

The equation $\frac{x^{2}}{2-\lambda}-\frac{y^{2}}{\lambda-5}-1=0$ represents an ellipse,if

If the angle between the lines joining the end points of the minor axis of an ellipse with its foci is $\frac{\pi}{2}$,then the eccentricity of the ellipse is

Find the equation of the ellipse whose axes are the coordinate axes,which passes through the point $(-3, 1)$ and has an eccentricity of $\sqrt{2/5}$.

$S^{\prime}$ is the focus of the ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1, (b < 5)$ lying on the negative $X$-axis and $P(\theta)$ is a point on this ellipse. If the distance between the foci of this ellipse is $8$ and $S^{\prime}P = 7$,then $\theta =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module