The locus of the midpoints of the portion of | vedclass

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

Question

(C) Given the ellipse $\frac{x^2}{2}+\frac{y^2}{1}=1$,where $a^2=2$ and $b^2=1$.The equation of the tangent at any point $(\sqrt{2}\cos 	heta, \sin \

Vedclass · Accepted Answer

$\frac{1}{2x^2}+\frac{1}{4y^2}=1$

Vedclass · Answer

(C) Given the ellipse $\frac{x^2}{2}+\frac{y^2}{1}=1$,where $a^2=2$ and $b^2=1$.The equation of the tangent at any point $(\sqrt{2}\cos 	heta, \sin 	heta)$ is given by $\frac{x \cos 	heta}{\sqrt{2}} + y \sin 	heta = 1$.The tangent intersects the $x$-axis at point $A$ where $y=0$,so $A = (\frac{\sqrt{2}}{\cos 	heta}, 0)$.The tangent intersects the $y$-axis at point $B$ where $x=0$,so $B = (0, \frac{1}{\sin 	heta})$.Let $(h, k)$ be the midpoint of the segment $AB$. Then $h = \frac{\sqrt{2}}{2 \cos 	heta}$ and $k = \frac{1}{2 \sin 	heta}$.From these,we have $\cos 	heta = \frac{\sqrt{2}}{2h} = \frac{1}{\sqrt{2}h}$ and $\sin 	heta = \frac{1}{2k}$.Using the identity $\sin^2 	heta + \cos^2 	heta = 1$,we get $(\frac{1}{2k})^2 + (\frac{1}{\sqrt{2}h})^2 = 1$.This simplifies to $\frac{1}{4k^2} + \frac{1}{2h^2} = 1$.Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

The locus of the midpoints of the portion of the tangents of the ellipse $\frac{x^2}{2}+\frac{y^2}{1}=1$ intercepted between the coordinate axes is

Similar Questions

With the origin as a focus and $x = 4$ as the corresponding directrix, a family of ellipses is drawn. Then the locus of an end of the minor axis is

If the distance between the two foci of an ellipse is equal to its minor axis,then the eccentricity of the ellipse is:

The equation of the normal to the curve $4x^2 + 9y^2 = 36$ at the point where the parametric angle is $\theta = \frac{7\pi}{4}$ is

If the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ meets the line $\frac{x}{7}+\frac{y}{2\sqrt{6}}=1$ on the $x$-axis and the line $\frac{x}{7}-\frac{y}{2\sqrt{6}}=1$ on the $y$-axis,then the eccentricity of the ellipse is

The area of the quadrilateral formed by drawing tangents at the ends of the latus recta of the ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module