A variable tangent to the ellipse frac{x^2}{a | vedclass

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

Question

(D) The equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.The equation of the tangent at point $(a \cos 	heta, b \sin 	heta)$ is $\frac

Vedclass · Accepted Answer

$\frac{a^2}{x^2}+\frac{b^2}{y^2}=4$

Vedclass · Answer

(D) The equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.The equation of the tangent at point $(a \cos 	heta, b \sin 	heta)$ is $\frac{x \cos 	heta}{a}+\frac{y \sin 	heta}{b}=1$.The tangent meets the $x$-axis at $A\left(\frac{a}{\cos 	heta}, 0ight)$ and the $y$-axis at $B\left(0, \frac{b}{\sin 	heta}ight)$.Let $(x, y)$ be the midpoint of $AB$. Then:$x = \frac{a}{2 \cos 	heta} \Rightarrow \cos 	heta = \frac{a}{2x}$$y = \frac{b}{2 \sin 	heta} \Rightarrow \sin 	heta = \frac{b}{2y}$Using the identity $\sin^2 	heta + \cos^2 	heta = 1$,we get:$\left(\frac{b}{2y}ight)^2 + \left(\frac{a}{2x}ight)^2 = 1$$\frac{b^2}{4y^2} + \frac{a^2}{4x^2} = 1$$\frac{a^2}{x^2} + \frac{b^2}{y^2} = 4$.Thus,the locus of the midpoint is $\frac{a^2}{x^2} + \frac{b^2}{y^2} = 4$.

$A$ variable tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ makes intercepts on both the axes. The locus of the midpoint of the portion of the tangent between the coordinate axes is

Similar Questions

The equation $\sqrt{(x - 2)^2 + y^2} + \sqrt{(x + 2)^2 + y^2} = 4$ represents:

Let the length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b)$ be $30$. If its eccentricity is the maximum value of the function $f(t)=-\frac{3}{4}+2t-t^{2}$, then $(a^{2}+b^{2})$ is equal to -

The tangent to the ellipse $9x^2 + 16y^2 = 288$ making equal intercepts on the co-ordinate axes intersects the $X$-axis and the $Y$-axis in the points $A$ and $B$ respectively. Then $\text{Area}(\triangle OAB) = $ (where $O$ is origin)

An ellipse inscribed in a semi-circle touches the circular arc at two distinct points and also touches the bounding diameter. Its major axis is parallel to the bounding diameter. When the ellipse has the maximum possible area,its eccentricity is

$A$ line passing through the endpoint of the major axis $A$ and the endpoint of the minor axis $B$ of an ellipse $\frac{x^2}{9} + y^2 = 1$ touches its auxiliary circle at point $M$. Find the area of the triangle with vertices $A, M,$ and the origin $O$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module