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

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.

• What did you get for the slope of the tangent line? Sep 30 '13 at 22:21

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 $(a, b) = (-1,1)$:

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

• This needs a TU! +1 Sep 29 '13 at 23:02

Hint: If $L_1$ and $L_2$ are two perpendicular lines in the plane with non-zero slope, then the product of their slopes is $-1$.

If you are able to find the tangent line, then in particular you can find the slope of the tangent line. You can use this to find the slope of the normal. You also know a point on the normal line (namely $(-1,1)$), so you can find the equation of your line using point-slope form.

The curve is a degenerate conic that is equivalent to a pair of straights:

$$(x+y)(7x-10y)=0$$,

therefore the problem leads to finding the perpendicular to the bisector of the second and fourth quadrants of Cartesian plane, passing through the point of the degenerate curve of coordinates \$(-1,1). The reasoning of "amWhy" above is valid and therefore the line has equation:

$$y=x+2.$$