Where does the normal line intersect the second time?

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SUMMARY

The discussion focuses on finding the second intersection point of the normal line to the ellipse defined by the equation x² - xy + y² = 3 at the point (-1, 1). The derivative was calculated as y' = (2x - y) / (x - 2y), yielding a slope of 1 at the given point. Consequently, the slope of the normal line is -1, leading to the equation y = -x. The second intersection point with the ellipse is confirmed to be (1, -1).

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Where does the normal line to the ellipse x2-xy+y2=3 at the point (-1,1) intersect the ellipse a second time?So I took the derivative of the equation to get:
y'=(2x-y)/(x-2y)
Then I put (-1,1) into the equation, to get a slope of 1.
So, for the normal line I got a slope of -1, which has the equation y=-x.

And that's as far as I got, I know the answer is suppose to be (1,-1).
 
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Nevermind, I figured it out.
 

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