The locus of the midpoint of the line segment | vedclass

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

Question

(C) Let $P = (4, 3)$ and $Q = (2\cos	heta, \sqrt{2}\sin	heta)$ be a point on the ellipse $x^{2} + 2y^{2} = 4$.Let $D(h, k)$ be the midpoint of $PQ$.

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(C) Let $P = (4, 3)$ and $Q = (2\cos	heta, \sqrt{2}\sin	heta)$ be a point on the ellipse $x^{2} + 2y^{2} = 4$.Let $D(h, k)$ be the midpoint of $PQ$.Then,$h = \frac{4 + 2\cos	heta}{2} = 2 + \cos	heta \implies \cos	heta = h - 2$.And $k = \frac{3 + \sqrt{2}\sin	heta}{2} \implies \sqrt{2}\sin	heta = 2k - 3 \implies \sin	heta = \frac{2k - 3}{\sqrt{2}}$.Using the identity $\cos^{2}	heta + \sin^{2}	heta = 1$,we get:$(h - 2)^{2} + \left(\frac{2k - 3}{\sqrt{2}}ight)^{2} = 1$$(h - 2)^{2} + \frac{4(k - 1.5)^{2}}{2} = 1$$(h - 2)^{2} + 2(k - 1.5)^{2} = 1$Dividing by $1$,we get $\frac{(h - 2)^{2}}{1} + \frac{(k - 1.5)^{2}}{1/2} = 1$.This is an ellipse with $a^{2} = 1$ and $b^{2} = 1/2$.Since $a^{2} > b^{2}$,the eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{1/2}{1}} = \sqrt{1 - 1/2} = \sqrt{1/2} = \frac{1}{\sqrt{2}}$.

The locus of the midpoint of the line segment joining the point $(4,3)$ and the points on the ellipse $x^{2}+2y^{2}=4$ is an ellipse with eccentricity:

Similar Questions

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 $\theta$ is the eccentric angle of a point on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ such that the distance of the point from the center is $5$, then $\theta = ......$

Let $P$ and $Q$ be the foci of an ellipse and let $R$ be one end of its minor axis. If $\triangle PQR$ is an equilateral triangle,then the eccentricity of the ellipse is equal to

If the line $x - 2y = 12$ is tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(3, -4.5)$,then the length of the latus rectum of the ellipse is:

The curve represented by $x = 3(\cos t + \sin t)$ and $y = 4(\cos t - \sin t)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module