Are you sure you've got the slope of the line correct? If you differentiate the equation then you get . Put x=-1 and y=1 and you find . So the perpendicular line has slope -1.
Helping out my brother in law with a Calc problem and I havn't taken calc in 8 years or so, and I need a little help with a basic problem. Hopefully you guys can check my work.
Problem:
Where does the line perpendicular to the ellipse
x^2 - xy + y^2 = 3
at the point (-1,1) intersect the ellipse a second time?
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Without getting nuts, I found the line equation for the perpendicular line to be
y=(1/3)x + (4/3)
Thew this back into the elliptical equation, boiled it down to the quadratic equation 7x^2 - 4x - 11 = 0 with the roots -1 and 11/7. Taking 11/7 into the line equation for the perpendicular line I get the value for y as 13/7.
So my answer is (11/7 , 13/7) as the second point.
Can someone verify this for me?