Find the points on the curve `5x^2-8x y+5y^2=4` whose distance from the origin is maximum or minimum.

Explanation

Let`(r,theta)` be the polar coordinates of any point P on the where r is the distance of the point from the origin .Then
`r^(2)[(5(cos^(2)theta+sin^(2)theta)-8 sin theta cos theta)]=4`
`r^(2)=(4)/(5-4sin 2theta)`
`r^(2)` is maximum when 5-4 sin 2 `theta` is minumum i.e 5-4=1
(When sin 2`theta` =1) or
`2theta =90^(@) "or" theta=45^(@) "or" = pm 2, theta 45^(@)`
Ag in `r^(2)` is minimum when 5-4 sin 2 `theta` is maximum i.e
5+4=9 when
sin `2theta=-1 "or" 2theta =(3pi)/(2) "or" theta=(3pi)/(4)`
or `r=pm (2)/(33),theta=(3pi)/(4)`
Hence the points are `(r cos theta, r sin theta)` where r and `theta` are given by equation S(1) and (2)
Thus we get four points `sqrt(2).sqrt(2), sqrt(-2), -sqrt(2)`,
`sqrt(3)/(3),-sqrt(2)/(3) "and" sqrt(2)/(3),sqrt(2)/(3)`