Question

The locus of the point represented by
x = 3(cost+sint), y = 2(cost-sint), is

​

Answer 1

$\textbf{Given:}$

$x=3(cost+sint)$

$y=2(cost-sint)$

$\textbf{To find:}$

$\text{The locus of the moving point}$

$\textbf{Solution:}$

$\text{Consider,}$

$x=3(cost+sint)$

$\dfrac{x}{3}=cost+sint$ .......(1)

$y=2(cost-sint)$

$\dfrac{y}{2}=cost-sint$ .......(2)

$\text{Squaring and adding equations (1) and (2)}$

$\dfrac{x^2}{9}+\dfrac{y^2}{4}=(cost+sint)^2+(cost-sint)^2$

$\dfrac{x^2}{9}+\dfrac{y^2}{4}=cos^2t+sin^2t+2\,cost\,sint+cos^2t+sin^2t-2\,cost\,sint$

$\dfrac{x^2}{9}+\dfrac{y^2}{4}=(cos^2t+sin^2t)+(cos^2t+sin^2t)$

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

$\dfrac{x^2}{9}+\dfrac{y^2}{4}=2$

$\implies\dfrac{x^2}{18}+\dfrac{y^2}{8}=1$

$\textbf{Answer:}$

$\textbf{The locus of the given point is \bf\dfrac{x^2}{18}+\dfrac{y^2}{8}=1 }$

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