Question

The number of points on the curve $5x^2-8xy+5y^2=4$ whose distance from the origin is maximum or minimum is ________.

Answer 1

$5x^2-8xy+5y^2=4$

$(4x^2-4xy+y^2)+(x^2-4xy+4y^2)=4$

$(2x-y{)}^2+(x-2y{)}^2=4$

$\frac{2x-y{)}^2}{4}+\frac{(x-2y{)}^2}{4}=1$

From this equation we can see that the major

axis of the ellipse is $(2x-y=0)$ and the minor axis $(x-2y=0)$

two axis intersect at origin thus centre of ellipse is at origin

thus point doest to origin on the ellipse the point where the minor axis and ellipse intersect

Similarly

the point farthest to origin on the ellipse on the ellipse the point where the major axis acid ellipse intersect

Let the point lying on major axis be $(a,b)$

$(a,b)$ also lies on line $2x-y=0$

$2a-b=0$

$a=\frac{b}{2}$ thus the point $(\frac{b}{2},b)$

also lies on ellipse

$(b-b{)}_4^2+{\left(\frac{\frac{b}{2}-2b}{4}\right)}^2=1$

$\frac{9b^2}{4}=4\Rightarrow b=\pm \frac{4}{3}$

Hence the point $(a,b)$ is either

$(\frac{2}{3},\frac{4}{3})$ or $(\frac{-2}{3},\frac{-4}{3})$ Answer

Let the point lying on minor axis be $(c,d)$ this point lies on line $x=2y$

$\therefore c=2d$

hence point becomes $(2d,d)$

the point also satisfy equation of

$\frac{(4d-d{)}^2}{4}+\frac{(2d-2d{)}^2}{4}=1$

$\frac{3d^2}{4}=1$

$d=\pm \frac{2}{3}$

thus the point $(c,d)$ is

either $(\frac{4}{3},\frac{2}{3})$ or $(\frac{-4}{3},\frac{-2}{3})$