Find the condition for the line ax + by + c = | vedclass

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

Question

(C) Let a point $P(2 \cos 	heta, 6 \sin 	heta)$ be on the ellipse $\frac{x^2}{4} + \frac{y^2}{36} = 1$. The equation of the normal to the ellipse at

Vedclass · Accepted Answer

$\frac{1}{a^2} + \frac{9}{b^2} = \frac{256}{c^2}$

Vedclass · Answer

(C) Let a point $P(2 \cos 	heta, 6 \sin 	heta)$ be on the ellipse $\frac{x^2}{4} + \frac{y^2}{36} = 1$. The equation of the normal to the ellipse at point $P$ is given by $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$. Here $a^2 = 4$ and $b^2 = 36$. Substituting the coordinates $(2 \cos 	heta, 6 \sin 	heta)$,we get: $\frac{4x}{2 \cos 	heta} - \frac{36y}{6 \sin 	heta} = 4 - 36$. Simplifying,we have $2x \sec 	heta - 6y \operatorname{cosec} 	heta = -32$,or $2x \sec 	heta - 6y \operatorname{cosec} 	heta + 32 = 0$. Comparing this with $ax + by + c = 0$,we have $\frac{a}{2 \sec 	heta} = \frac{b}{-6 \operatorname{cosec} 	heta} = \frac{c}{32}$. This gives $\cos 	heta = \frac{2c}{32a} = \frac{c}{16a}$ and $\sin 	heta = -\frac{6c}{32b} = -\frac{3c}{16b}$. Using $\cos^2 	heta + \sin^2 	heta = 1$,we get $\frac{c^2}{256a^2} + \frac{9c^2}{256b^2} = 1$,which simplifies to $\frac{1}{a^2} + \frac{9}{b^2} = \frac{256}{c^2}$.

Find the condition for the line $ax + by + c = 0$ to be a normal to an ellipse $\frac{x^2}{4} + \frac{y^2}{36} = 1$.

Similar Questions

Tangents are drawn to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$ at the ends of both latus recta. The area of the quadrilateral so formed is

If the length of the minor axis of an ellipse is equal to one-fourth of the distance between the foci,then the eccentricity of the ellipse is:

$P$ is a point on the conic $a^2 x^2+b^2 y^2=a^2(a^2+b^2-y^2)$ and $S$ is a focus of that conic. $M$ is the foot of the perpendicular from $P$ onto a directrix of that conic nearer to $S$. If $PM = K SP$,then $K=$

Let $T_1$ be the tangent drawn at a point $P(\sqrt{2}, \sqrt{3})$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{6}=1$. If $(\alpha, \beta)$ is the point where $T_1$ intersects another tangent $T_2$ to the ellipse perpendicularly,then $\alpha^2+\beta^2=$

Let $P$ be an arbitrary point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a > b > 0$. Suppose $F_1$ and $F_2$ are the foci of the ellipse. The locus of the centroid of the $\triangle P F_1 F_2$ as $P$ moves on the ellipse is

Vedclass Test Series

Exam Paper Generator

Online Exam Module