The centre of a circle C is at the centre of | vedclass

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

Question

(B) Let the distance between the foci $F_1$ and $F_2$ be $2c$. Since the circle $C$ is centered at the origin and passes through the foci $(\pm c, 0)$

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) Let the distance between the foci $F_1$ and $F_2$ be $2c$. Since the circle $C$ is centered at the origin and passes through the foci $(\pm c, 0)$,its radius is $r = c$.Since $P$ lies on the circle $C$,the distance $OP = c$. Also,since $F_1$ and $F_2$ are on the circle,the angle $\angle F_1PF_2 = 90^\circ$ because $F_1F_2$ is the diameter of the circle.Let $PF_1 = x$ and $PF_2 = y$. Since $P$ lies on the ellipse,by the definition of an ellipse,$x + y = 2a = 17$.In the right-angled triangle $PF_1F_2$,the area is $\frac{1}{2} xy = 30$,so $xy = 60$.We know that $(x + y)^2 = x^2 + y^2 + 2xy$. By the Pythagorean theorem in $	riangle PF_1F_2$,$x^2 + y^2 = (F_1F_2)^2 = (2c)^2 = 4c^2$.Thus,$(2a)^2 = 4c^2 + 2xy$,which gives $17^2 = 4c^2 + 2(60)$.$289 = 4c^2 + 120$.$4c^2 = 169$.$2c = \sqrt{169} = 13$.The distance between the foci is $2c = 13$.

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 area of the greatest rectangle that can be inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

An ellipse passing through $(4 \sqrt{2}, 2 \sqrt{6})$ has foci at $(-4, 0)$ and $(4, 0)$. Then,its eccentricity is

The eccentric angles of the extremities of the latus recta of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are given by

The length of the latus rectum of an ellipse is $6$ units and the distance between a focus and its nearest vertex on the major axis is $\frac{5}{3}$ units. If $e$ is the eccentricity of this ellipse,then $e$ satisfies the equation:

Vedclass Test Series

Exam Paper Generator

Online Exam Module