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Ellipse

Section: Mathematics

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Question : 68 of 117

Marks: +1, -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

[31 Aug 2021 Shift 2]

$9 x^{2}+4 y^{2}+18 x+8 y+145=0$
$36 x^{2}+16 y^{2}+90 x+56 y+145=0$
$36 x^{2}+16 y^{2}+108 x+80 y+145=0$
$36 x^{2}+16 y^{2}+72 x+32 y+145=0$

Solution:

Let

(2 \sin \theta, 3 \cos \theta)

be the point on ellipse

\frac{x^{2}}{4}+\frac{y^{2}}{9}=1

and let mid -point of the line segments joining

(-3,-5)

and

(2 \sin \theta, 3 \cos \theta)

will be

(h, k)

.

Then,

\frac{2 \sin \theta-3}{2}=h, \frac{3 \cos \theta-5}{2}=k

\Rightarrow \sin \theta=\frac{2 h+3}{2}, \cos \theta=\frac{2 k+5}{3}

\therefore \sin ^{2} \theta+\cos ^{2} \theta=1

\Rightarrow \left(\frac{2 h+3}{2}\right)^{2}+\left(\frac{2 k+5}{3}\right)^{2}=1

\Rightarrow \frac{4 h^{2}+9+12 h}{4}+\frac{4 k^{2}+25+20 k}{9}=1

36 h^{2}+16 k^{2}+108 h+80 k+145=0

LOCUS of

(h, k)

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

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