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10-2. Parabola, Ellipse, Hyperbola
hard
The locus of the mid point of the line segment joining the point $(4,3)$ and the points on the ellipse $x^{2}+2 y^{2}=4$ is an ellipse with eccentricity
A
$\frac{\sqrt{3}}{2}$
B
$\frac{1}{2 \sqrt{2}}$
C
$\frac{1}{\sqrt{2}}$
D
$\frac{1}{2}$
(JEE MAIN-2022)
Solution
$\frac{ x ^{2}}{4}+\frac{ y ^{2}}{2}=1$
Coordinate of $D$ is
$\left(\frac{2 \cos \theta+4}{2}, \frac{\sqrt{2} \sin \theta+3}{2}\right) \equiv(h, k)$
$\frac{2 h -4}{2}=\cos \theta$
$\frac{2 k -3}{\sqrt{2}}=\sin \theta$
$(i)^{2}+(\text { ii })^{2}$, then we get
$\left(\frac{2 h -4}{2}\right)^{2}+\left(\frac{2 k -3}{\sqrt{2}}\right)^{2}=1 \Rightarrow \frac{( x -2)^{2}}{1}+\frac{\left( y -\frac{3}{2}\right)^{2}}{\left(\frac{1}{2}\right)}=1$
$\therefore \quad$ Required eccentricity is
$e =\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$
Standard 11
Mathematics
