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 :
Solution
<p>Let $P(2\cos \theta ,\,\sqrt 2 \sin \theta )$ be any point on ellipse ${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$ and Q(4, 3) and let (h, k) be the mid point of PQ</p>
<p>then $h = {{2\cos \theta + 4} \over 2},\,k = {{\sqrt 2 \sin \theta + 3} \over 2}$</p>
<p>$\therefore$ $\cos \theta = h - 2,\,\sin \theta = {{2k - 3} \over {\sqrt 2 }}$</p>
<p>$\therefore$ ${(h - 2)^2} + {\left( {{{2k - 3} \over {\sqrt 2 }}} \right)^2} = 1$</p>
<p>$$ \Rightarrow {{{{(x - 2)}^2}} \over 1} + {{{{\left( {y - {3 \over 2}} \right)}^2}} \over {{1 \over 2}}} = 1$$</p>
<p>$\therefore$ $e = \sqrt {1 - {1 \over 2}} = {1 \over {\sqrt 2 }}$</p>
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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