If the length of the latus rectum of the elli | vedclass

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

Question

(D) The given equation is $x^{2} + 4y^{2} + 2x + 8y - \lambda = 0$.Completing the square,we get $(x^{2} + 2x + 1) + 4(y^{2} + 2y + 1) = \lambda + 1 +

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(D) The given equation is $x^{2} + 4y^{2} + 2x + 8y - \lambda = 0$.Completing the square,we get $(x^{2} + 2x + 1) + 4(y^{2} + 2y + 1) = \lambda + 1 + 4$.$(x + 1)^{2} + 4(y + 1)^{2} = \lambda + 5$.Dividing by $\lambda + 5$,we get $\frac{(x + 1)^{2}}{\lambda + 5} + \frac{(y + 1)^{2}}{(\lambda + 5)/4} = 1$.Here,$a^{2} = \lambda + 5$ and $b^{2} = \frac{\lambda + 5}{4}$,so $a = \sqrt{\lambda + 5}$ and $b = \frac{\sqrt{\lambda + 5}}{2}$.The length of the latus rectum is $\frac{2b^{2}}{a} = 4$.Substituting the values,$\frac{2(\lambda + 5)/4}{\sqrt{\lambda + 5}} = 4$.$\frac{\lambda + 5}{2\sqrt{\lambda + 5}} = 4 \implies \frac{\sqrt{\lambda + 5}}{2} = 4 \implies \sqrt{\lambda + 5} = 8$.Squaring both sides,$\lambda + 5 = 64$,so $\lambda = 59$.The length of the major axis $l = 2a = 2\sqrt{\lambda + 5} = 2(8) = 16$.Therefore,$\lambda + l = 59 + 16 = 75$.

If the length of the latus rectum of the ellipse $x^{2} + 4y^{2} + 2x + 8y - \lambda = 0$ is $4$,and $l$ is the length of its major axis,then $\lambda + l$ is equal to $......$

Similar Questions

On the ellipse $4x^{2} + 9y^{2} = 1$, the points at which the tangents are parallel to the line $8x = 9y$ are:

If the area of the auxiliary circle of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ (where $a > b$) is twice the area of the ellipse,then the eccentricity of the ellipse is

The locus of a point such that the sum of its distances from the points $(0, 2)$ and $(0, -2)$ is $6$ is:

The line $x \cos \alpha + y \sin \alpha = p$ will be a tangent to the conic $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,if

If $4x + 2y + n = 0$ is a normal to the ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$,then $n = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module