Find the equation of the tangent line to the curve

Question

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.

Answer 1

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) =nx^n-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.

Comments

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

and m  = 9/13 and b  = - 40/13.

Answer 2

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 .

 

Answer 3

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 .