An ellipse frac{(x-x_0)^2}{a^2} + frac{(y-y_0 | vedclass

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

Question

(C) Since the ellipse is tangent to both axes in the first quadrant,its center is $(a, b)$. The foci are $(a \pm ae, b)$ or $(a, b \pm ae)$. Given the

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) Since the ellipse is tangent to both axes in the first quadrant,its center is $(a, b)$. The foci are $(a \pm ae, b)$ or $(a, b \pm ae)$. Given the geometry,the center is $(a, b)$ and the distance from the origin to the center is $\sqrt{a^2 + b^2}$.For an ellipse tangent to the axes,$x_0 = a$ and $y_0 = b$. The foci are $(a \pm ae, b)$.Given $	riangle OF_1F_2$ is isosceles with $\angle OF_1F_2 = 120^{\circ}$,we have $OF_1 = F_1F_2 = 2ae$.Using the law of cosines in $	riangle OF_1F_2$ or coordinate geometry properties,we find $e = \frac{1}{2}$.

Similar Questions

The locus of the midpoints of the intercepted portion of the tangents by the coordinate axes,which are drawn to the ellipse $x^2+2y^2=2$,is

The major and minor axes of an ellipse are along the $X$-axis and $Y$-axis respectively. If its latus rectum is of length $4$ and the distance between the foci is $4 \sqrt{2}$,then the equation of that ellipse is

If $A$ and $B$ are two fixed points and $P$ is a variable point such that $PA + PB = 4$,then the locus of $P$ is a/an

An ellipse having foci at $(3, 3)$ and $(-4, 4)$ and passing through the origin has eccentricity equal to

If the normal drawn at the point $P(\frac{\pi}{4})$ on the ellipse $x^2+4y^2-4=0$ meets the ellipse again at $Q(\alpha, \beta)$,then $\alpha=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module