# Find the equation of the tangent line to the curve

2(x^2+y^2)^2 = 25(x^2-y^2) at the point ( 3 , -1 ). The equation of this tangent line can be written in the form y = mx+b. Find m and b.

asked Mar 30, 2013 in CALCULUS

2(x2 + y2)2 =25(x2 - y2)

we can find the slope

Recall : (a + b)2 = a2 +2ab + b2

2(x4 +2x2y2 + y4) = 25(x2 - y2)

2x4 + 4x2y2 + 2y4 = 25x2 - 25y2

Derivative with respective x  to each side

Derivative power rule d / dx(xn) =nxn-1 and derivative product rule d / dx(u v) = u v1 + vu1

2(4x3) + 4(2xy2 + 2ydy / dx x2) + 4y3 dy / dx = 25(2x) - 25(2ydy / dx)

8x3 + 8xy2 + 8ydy / dx x2 + 4y3 dy /dx  = 50x - 50ydy / dx

We can rewrite the term coefficient  of dy / dx and rewrite the other coefficients

Add 50ydy / dx to each side

8x3 + 8xy2 + 8ydy / dx x2 + 4y3 dy / dx + 50y dy / dx = 50x - 50ydy / dx + 50ydy / dx

8x3 + 8xy2 + 8yx2dy / dx + 4y3 dy / dx + 50y dy / dx = 50x

Subtract 8x3 + 8xy2 from each side

8x3 + 8xy2 + 8ydy / dx x2 + 4y3 dy / dx + 50y dy / dx  - (8x3 + 8xy2)= 50x - (8x3 + 8xy2)

Simplify

8ydy / dx x2 + 4y3 dy / dx + 50y dy / dx = 50x - 8x3 - 8xy2

Take  out common term dy / dx

dy / dx(8yx2 + 4y3 + 50y) = 50x - 8x3 - 8xy2

Take out common term 2y  in the L.H.S and Take out common term 2x in the R.H.S

dy / dx 2y(4x2 + 2y2 + 25) = 2x(25 - 4x2 -4y2)

Divide each side by 2y(4x2 + 2y2 + 25)

There fore dy /dx = 2x(4x2 + 2y2 + 25) / 2y(4x2 + 2y2 + 25)

dy / dx = x   (4x2 + 2y2 +25) / y(4x2 + 2y2 + 25)

Substitute x =3 and y = -1 in the dy / dx

dy /dx = 3(4(9) + 2(1) + 25) / -1(4(9) +2(1) + 25

m =  dy / dx = -3(36 +27) / (36 + 27)

m = -3

There fore the slope m = -3 and thepoint (3 , -1) then the tangent line equation y - y1 =m(x - x1)

y -(-1) = -3(x - 3)

y + 1 = -3x + 9

Subtract 1 to each side

y = -3x + 8

This is form a y = mx + b

There fore m = -3 and b = 8.

edited Apr 1, 2013

Tangent line of  2(x2 + y2)2 =25(x2 - y2) at the point (3, -1) is .

and = 9/13 and b  = - 40/13.

The curve equation is

The tangent to a curve is a straight line that touches the curve at a certain point

and has exactly the same slope as the curve at that point.

The derivative of a function is the slope of the function at a certain point,and so the tangent line to the curve.

Apply formula :

• Find the derivative of this function.

Differentiating on each side with respect of x .

Continuous....

Curve and tangent line passes through the point (3, - 1)

• Substitute the (x,y ) value of this point into the derived function .

This is the slope of tangent line to the curve at (3, - 1).

• To find the tangent line equation, substitute the values of m = 9/13 and (x, y ) = (3, - 1) in the slope intercept form of an equation.

Therfore m = 9/13 and b = - 40/13.

Tangent line equation is .