A normal line to a curve passes through a point P on the curve perpendicular to the line tangent to the curve at P. Use the following equation and graph to determine an equation of the normal line at the given point and illustrate your work by graphing the curve with the normal line.

6x^2 + 4xy + 3y^2 = 117; (3,3)

Question

A normal line to a curve passes through a point P on the curve perpendicular to the line tangent to the curve at P. Use the following equation and graph to determine an equation of the normal line at the given point and illustrate your work by graphing the curve with the normal line. 

6x^2 + 4xy + 3y^2 = 117; (3,3)

coconutjj · Answer

hint: dy/dx

anonymous · Answer

my answer was y = 5/8x + 39/8 but it was wrong

anonymous · Answer

@CoconutJJ

michele_laino · Answer

please keep in mind that the slope m of your line, is:
$$m =  - \frac{1}{{y'}}$$

michele_laino · Answer

more precisely:
$$m =  - \frac{1}{{y'\left( {3,3} \right)}}$$