A curve has the equation, 6x^2 +3xy−y^2 +6=0 and passes through the point A (-5, 10). Find the equation of the normal to the curve at A.
Use implicit differentiation on original equation-
12x + 3x(dy/dx) + 3y - 2y(dy/dx) = 0
dy/dx= -12x -3y/(3x-2y) at A, x= -5 and y= 10 therefore, dy/dx=-6/7
To find the normal of the curve, use the negative reciprocal of the gradient calculated for the gradient of the normal, 7/6. Now we can form the equation,
y= mx + c
y= 7/6.x + c
Input y and x values in the question to calculate c;
c= 10-7/6(-5)= 95/6
So, y= 7/6x + 95/6
CG
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