Let the ellipse 3x^2 + py^2 = 4 pass through | vedclass

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

Question

(C) The equation of the circle is $x^2 + y^2 - 2x - 4y - 11 = 0$. The centre $C$ is $(1, 2)$ and the radius $r = \sqrt{1^2 + 2^2 - (-11)} = \sqrt{1 +

Vedclass · Accepted Answer

$70$

Vedclass · Answer

(C) The equation of the circle is $x^2 + y^2 - 2x - 4y - 11 = 0$. The centre $C$ is $(1, 2)$ and the radius $r = \sqrt{1^2 + 2^2 - (-11)} = \sqrt{1 + 4 + 11} = 4$. Since the ellipse $3x^2 + py^2 = 4$ passes through $(1, 2)$,we have $3(1)^2 + p(2)^2 = 4$,which gives $3 + 4p = 4$,so $p = \frac{1}{4}$. The ellipse equation is $3x^2 + \frac{1}{4}y^2 = 4$,or $\frac{x^2}{4/3} + \frac{y^2}{16} = 1$. Here $a^2 = \frac{4}{3}$ and $b^2 = 16$. Since $b^2 > a^2$,this is a vertical ellipse. The eccentricity $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{4/3}{16}} = \sqrt{1 - \frac{1}{12}} = \sqrt{\frac{11}{12}}$. The focal distances of a point $(x_0, y_0)$ on an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are $b \pm ey_0$. Thus,$f_1 = b + ey_0 = 4 + \sqrt{\frac{11}{12}} 	imes 2$ and $f_2 = b - ey_0 = 4 - \sqrt{\frac{11}{12}} 	imes 2$. Then $f_1 f_2 = (4)^2 - (\sqrt{\frac{11}{12}} 	imes 2)^2 = 16 - (\frac{11}{12} 	imes 4) = 16 - \frac{11}{3} = \frac{48 - 11}{3} = \frac{37}{3}$. Finally,$6f_1f_2 - r = 6(\frac{37}{3}) - 4 = 2(37) - 4 = 74 - 4 = 70$.

Let the ellipse $3x^2 + py^2 = 4$ pass through the centre $C$ of the circle $x^2 + y^2 - 2x - 4y - 11 = 0$ with radius $r$. Let $f_1, f_2$ be the focal distances of the point $C$ on the ellipse. Then $6f_1f_2 - r$ is equal to

Similar Questions

The locus of the point of intersection of two perpendicular tangents to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is:

$A$ tangent is drawn to the ellipse $\frac{x^2}{32} + \frac{y^2}{8} = 1$ from the point $A(8, 0)$ to touch the ellipse at point $P$. If the normal at $P$ meets the major axis of the ellipse at point $B$,then the length $BC$ is equal to (where $C$ is the center of the ellipse) - ............ $units$

The eccentricity of the ellipse $4x^2 + y^2 - 8x + 2y + 1 = 0$ is

The equation of the tangent of the ellipse $4x^2 + 9y^2 = 36$ at the end of the latus rectum lying in the second quadrant is:

If tangents are drawn from the point $(2 + 13 \cos \theta, 3 + 13 \sin \theta)$ to the ellipse $\frac{(x-2)^2}{25} + \frac{(y-3)^2}{144} = 1$,then the angle between them is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module