Question

If $x,y\in R$, satisfies the equation
$\frac{(x-4{)}^2}{4}+\frac{y^2}{9}=1,$ then the difference between the largest and smallest value of the expression $\frac{x^2}{4}+\frac{y^2}{9}$ is

Answer 1

Parametric coordinates of any point on give ellipse will be of form
$x=2\cos \theta +4$ and $y=3\sin \theta$

Now $E=\frac{x^2}{4}+\frac{y^2}{9}$
$=\frac{(2\cos \theta +4{)}^2}{4}+{\sin }^2\theta$
$=\frac{4{\cos }^2\theta +16+16\cos \theta +4{\sin }^2\theta }{4}$
$=\frac{20+16\cos \theta }{4}$
$=5+4\cos \theta$
Hence $E_{max}-E_{min}=(9-1)=8$