[Math] Find the slope of the normal line to the curve $7x^2-10y^2=3xy$ at the point (-1,1).

Question

Find the slope of the normal line to the curve $7x^2-10y^2=3xy$ at the point (-1,1).

I can find tangent line pretty easy, find the normal line seems to get to me.

Please Help!!!

Answer 1

If the slope of the line tangent to the curve at the point $(-1, 1)$ is, let's call it $\;m,\;$ then the slope of the line normal to the curve at that point is perpendicular to the tangent line at $(-1)$ (the line "normal" to the curve at that point) is given by $$m' =\;-\dfrac 1m.\;$$

When you obtain that slope, use the point-slope form for which we need the value $m'$ and the point $(a, b) = (-1,1)$:

$$y - b = m'(x - a)$$