If 4x + 2y + n = 0 is a normal to the ellipse | vedclass

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

Question

(D) The equation of the line is $4x + 2y + n = 0$,which can be rewritten as $y = -2x - \frac{n}{2}$.Comparing this with $y = mx + c$,we get $m = -2$ a

Vedclass · Accepted Answer

$\pm 8$

Vedclass · Answer

(D) The equation of the line is $4x + 2y + n = 0$,which can be rewritten as $y = -2x - \frac{n}{2}$.Comparing this with $y = mx + c$,we get $m = -2$ and $c = -\frac{n}{2}$.The equation of the ellipse is $\frac{x^2}{36} + \frac{y^2}{16} = 1$,so $a^2 = 36$ and $b^2 = 16$.The condition for the line $y = mx + c$ to be a normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = \frac{(a^2 - b^2)^2 m^2}{a^2 + b^2 m^2}$.Substituting the values: $(-\frac{n}{2})^2 = \frac{(36 - 16)^2 (-2)^2}{36 + 16(-2)^2}$.$\frac{n^2}{4} = \frac{(20)^2 	imes 4}{36 + 16(4)} = \frac{400 	imes 4}{36 + 64} = \frac{1600}{100} = 16$.$n^2 = 16 	imes 4 = 64$.Therefore,$n = \pm 8$.

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

Similar Questions

The number of tangents to the circle $x^2 + y^2 = 3$ that are normal to the ellipse $4x^2 + 9y^2 = 36$ is

If a tangent having slope of $- \frac{4}{3}$ to the ellipse $\frac{x^2}{18} + \frac{y^2}{32} = 1$ intersects the major and minor axes in points $A$ and $B$ respectively,then the area of $\Delta OAB$ is equal to .................. $sq. \text{ units}$ ($O$ is the centre of the ellipse).

Find the length of the latus rectum of the ellipse $\frac{x^2}{36} + \frac{y^2}{49} = 1$.

$B$ is an extremity of the minor axis of an ellipse whose foci are $S$ and $S^{\prime}$. If $\angle SBS^{\prime}$ is a right angle, then the eccentricity of the ellipse is

What is the equation of the chord of the ellipse $\frac{x^2}{36} + \frac{y^2}{9} = 1$ that is bisected at the point $(2, 1)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module