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

Explanation

The figure
PC= b cosec `theta`
AP=a sec `theta`
AC=PC+AP
AC=b cosec `theta` + A sec `theta`
`d(AC)/(d theta)=- cosec theta cot theta cot theta + a sec theta tan theta`
`d(AC)/(d theta)=0`
or a sec `theta tan theta =b cosec theta cot theta`
or `tan theta =(b/a)^(1//3)`
Also `theta in (0,pi//2)`
`underset(theta rarr0)lim(a sec theta+b cosec theta)rarr 00`
and `underset(theta rarr)(pi-1)/(2)lim(a sec theta + b cosec theta)rarr 00`
Therefore, `theta = tan (-1)(b/a)^(1//3)` is a point of minima
For this value of `theta`
AC=`b sqrt(a^(2//3)+b^(2//3)/(b^(1//3))+A sqrt(a^(2//3)+b^(2//3)/(a^(1//3))`
`=sqrt(a^(2//3)+b^(2//3)b^(2//3)+a^(2//3))`
`=a^(2//3)+b^(2//3)^(3//2)`
Hence the minimum length of the hypotenuse is `(a^(2//3)+b^(2//3)+c^(3//2)`