Let the foci of the ellipse frac{x^{2}}{9}+y^ | vedclass

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

Question

(D) Given the ellipse equation is $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$. Comparing with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we have $a^{2}=9$ an

Vedclass · Accepted Answer

$x^{2}+y^{2}=8$

Vedclass · Answer

(D) Given the ellipse equation is $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$. Comparing with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we have $a^{2}=9$ and $b^{2}=1$. The eccentricity $e$ is given by $e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$. The foci are $(\pm ae, 0) = (\pm 3 	imes \frac{2\sqrt{2}}{3}, 0) = (\pm 2\sqrt{2}, 0)$. Let $P(h, k)$ be a point such that the foci $F_{1}(2\sqrt{2}, 0)$ and $F_{2}(-2\sqrt{2}, 0)$ subtend a right angle at $P$. Thus,the product of the slopes of $PF_{1}$ and $PF_{2}$ is $-1$. $\frac{k-0}{h-2\sqrt{2}} 	imes \frac{k-0}{h+2\sqrt{2}} = -1$. $\frac{k^{2}}{h^{2}-(2\sqrt{2})^{2}} = -1$. $k^{2} = -(h^{2}-8)$. $h^{2}+k^{2} = 8$. Replacing $(h, k)$ with $(x, y)$,the locus is $x^{2}+y^{2}=8$.

Let the foci of the ellipse $\frac{x^{2}}{9}+y^{2}=1$ subtend a right angle at a point $P$. Then,the locus of $P$ is

Similar Questions

If $3x + 4y = 12\sqrt{2}$ is a tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{9} = 1$ for some $a \in R$,then the distance between the foci of the ellipse is:

If tangents are drawn to the ellipse $x^2+2y^2=2$,then the locus of the mid-points of the intercepts made by those tangents between the coordinate axes is

The equation of the locus of a point $(2 \cos \theta-3, 3 \sin \theta-4)$ is

Let $F_1$ and $F_2$ be the foci of an ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$. $A$ ray from $F_1$ strikes the elliptical mirror at point $P$ and is reflected. What is the equation of the angle bisector of the angle between the incident ray and the reflected ray?

The length of the latus rectum of $x^2+3y^2=12$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module