Find the equation of the tangent and the normal lines to the curves x^2y+y^3x+3x-13=0 at the point (1,2)

Answer 1

To find the equation of the tangent line, we first need to find the derivative of the given curve. The equation of the curve is given by:

x^2y + y^3x + 3x - 13 = 0

Taking the derivative of both sides with respect to x:

2xy + x^2(dy/dx) + 3 + 3y^2x(dy/dx) = 0

Rearranging the terms:

(dy/dx)(x^2 + 3y^2x) = -2xy - 3

(dy/dx) = (-2xy - 3)/(x^2 + 3y^2x)

Now, to find the slope of the tangent line at the point (1,2), substitute x = 1 and y = 2 into the derivative:

(dy/dx) = (-2(1)(2) - 3)/(1^2 + 3(2)^2(1))

(dy/dx) = (-4 - 3)/(1 + 12)

(dy/dx) = -7/13

So the slope of the tangent line is -7/13.

To find the equation of the tangent line at the point (1,2), we can use the point-slope form:

y - y1 = m(x - x1)

where (x1, y1) = (1, 2) and m = -7/13:

y - 2 = (-7/13)(x - 1)

Multiply both sides by 13 to eliminate the fraction:

13y - 26 = -7(x - 1)

13y - 26 = -7x + 7

Rearrange the equation to get it in the standard form:

7x + 13y = 33

So the equation of the tangent line to the curve at the point (1,2) is 7x + 13y = 33.

To find the equation of the normal line, we need to find the negative reciprocal of the slope of the tangent line. The negative reciprocal of -7/13 is 13/7.

Using the same point-slope form as before, with m = 13/7:

y - 2 = (13/7)(x - 1)

Multiply both sides by 7 to eliminate the fraction:

7y - 14 = 13(x - 1)

7y - 14 = 13x - 13

Rearrange the equation to get it in the standard form:

13x - 7y = 1

So the equation of the normal line to the curve at the point (1,2) is 13x - 7y = 1.