The locus of mid-points of the line segments | vedclass

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

Question

(C) Let the point on the ellipse be $P(2 \cos 	heta, 3 \sin 	heta)$.Let the fixed point be $Q(-3, -5)$.Let $(h, k)$ be the mid-point of the segment

Vedclass · Accepted Answer

$36 x^{2}+16 y^{2}+108 x+80 y+145=0$

Vedclass · Answer

(C) Let the point on the ellipse be $P(2 \cos 	heta, 3 \sin 	heta)$.Let the fixed point be $Q(-3, -5)$.Let $(h, k)$ be the mid-point of the segment $PQ$. Then:$h = \frac{2 \cos 	heta - 3}{2} \implies 2 \cos 	heta = 2h + 3 \implies \cos 	heta = \frac{2h + 3}{2}$$k = \frac{3 \sin 	heta - 5}{2} \implies 3 \sin 	heta = 2k + 5 \implies \sin 	heta = \frac{2k + 5}{3}$Since $\cos^2 	heta + \sin^2 	heta = 1$,we have:$\left(\frac{2h + 3}{2}ight)^2 + \left(\frac{2k + 5}{3}ight)^2 = 1$$\frac{4h^2 + 12h + 9}{4} + \frac{4k^2 + 20k + 25}{9} = 1$Multiplying by $36$:$9(4h^2 + 12h + 9) + 4(4k^2 + 20k + 25) = 36$$36h^2 + 108h + 81 + 16k^2 + 80k + 100 = 36$$36h^2 + 16k^2 + 108h + 80k + 145 = 0$Replacing $(h, k)$ with $(x, y)$,the locus is $36x^2 + 16y^2 + 108x + 80y + 145 = 0$.

The locus of mid-points of the line segments joining $(-3,-5)$ and the points on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ is :

Similar Questions

If $\tan \theta_1 \cdot \tan \theta_2 = -\frac{a^2}{b^2}$,then the chord joining two points $\theta_1$ and $\theta_2$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ will subtend a right angle at:

One of the foci of an ellipse is $(2,-3)$ and its corresponding directrix is $2x+y=5$. If the eccentricity of the ellipse is $\frac{\sqrt{5}}{3}$,then the coordinates of the other focus are

For what value of $\lambda$ does the line $y = x + \lambda$ touch the ellipse $9x^2 + 16y^2 = 144$?

If a tangent having slope $m = \frac{1}{3}$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ is a normal to the circle $(x + 1)^2 + (y + 1)^2 = 1$,then $a^2$ lies in the interval:

Tangents are drawn from point $(1, 1)$ to the ellipse $S \equiv x^2 + 4y^2 - 2x + 8y + 1 = 0$. If $m_1, m_2$ $(m_1 > m_2)$ are the slopes of these tangents,then with respect to the given ellipse,the point $P(m_1, m_2)$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module