IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

sinjan.j

find the minimum dist b/w
x^2+y^2=9
and
curve 2x^2+10y^2+6xy=1

Conjurer

Here is a complete algebraic approach to the problem.

Let the x coordinate of any point lying on the curve be u.

Then , 2u^2 + 10y^2 +6uy =1

=> 5y^2 + 3uy +u^2-1/2 =0

y = -3 +- root(9u^2 -20u^2 +10)/10

y = -3 +- root(10-11u^2)/10

Now distance between centre of the circle and this curve is given by u^2 + (-3 +- root(10-11u^2)/10)^2

Appying AM=GM , when terms are equal we get,

u = 3 +- root(10-11u^2)/10

=> u^2 + 9 - 6y = 10-11y^2/100

Get u from here and put in distance equation. Then subtract the radius of the circle.

I hope its right.

Conjurer

My approach of applying AM = GM maybe wrong so if you dont get answer by that .Try squaring terms and make them zero.Basically you have to minimize u^2 + (-3 +- root(10-11u^2)/10)^2

sagarvaze

Minimum distance between the two conics when they have a common normal.
Find the normal from center of circle (origin) to given curve
minimum distance = length of normal - 3

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