[SOLVED] I forgot how to solve quadratic equations (again)

**billym** · December 1st 2009, 08:01 AM

Find the points on the curve $x^2 + 8xy + 7y^2 = 225$ that are closest to the origin.

Let $D^2=f(x,y)=x^2+y^2$ and $g(x,y)=x^2 + 8xy + 7y^2 - 225$

So: $\triangledown f=\lambda \triangledown g$

gives $\lambda = \frac{x}{x+4y}=\frac{y}{7y+4x} = 4x^2+6xy+4y^2 = 0$

Can somebody do this next step for me or remind what the method is for solving these...

**billym** · December 1st 2009, 08:03 AM

oops, calc error. nevermind.

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