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
To find the equation of the tangent line to the curve, we will use the concept of differentiation. The first step is to differentiate the equation of the curve with respect to x. Let's begin:
2(x^2 + y^2)^2 = 25(x^2 - y^2)
Differentiating both sides with respect to x:
d/dx[2(x^2+y^2)^2] = d/dx[25(x^2-y^2)]
To differentiate the left side, we will use the Chain Rule. Let's denote y as a function of x, y(x):
Chain Rule:
d/dx[f(g(x))] = f'(g(x)) * g'(x)
Applying the Chain Rule to the left side:
d/dx[2(x^2+y^2)^2] = 2 * d/dx[(x^2+y^2)^2] =
2 * 2(x^2+y^2)^1 * d/dx[(x^2+y^2)] =
4(x^2+y^2)(2x+2yy') ----(Eq. 1)
To differentiate the right side, we need to consider the two terms separately:
d/dx[25(x^2-y^2)] = 25 * d/dx[(x^2-y^2)] =
25 * (2x - 2yy') ----(Eq. 2)
Now we have the derivatives of both sides of the equation. Let's substitute the coordinates of the given point (3, 1) into these equations to find y':
From Eq. 1, substituting (3, 1):
4(3^2+1^2)(2(3)+2(1)y') = 4(10)(6+2yy')
120 = 80 + 80y'
40 = 80y'
y' = 40/80
y' = 1/2
From Eq. 2, substituting (3, 1):
25(2(3) - 2(1)y') = 25(6 - 2y')
50 - 50y' = 150 - 50y'
50y' = 100
y' = 100/50
y' = 2
We have now found the value of the derivative, y', which represents the slope of the tangent line at the point (3, 1).
Next, we will find the value of y at the point (3, 1) by substituting the coordinates into the original equation of the curve:
2(3^2 + y^2)^2 = 25(3^2 - y^2)
2(9 + y^2)^2 = 25(9 - y^2)
(9 + y^2)^2 = (9 - y^2)(25/2)
(9 + y^2)^2 = (225/2) - (25/2)y^2
Substituting y = 1:
(9 + 1^2)^2 = (225/2) - (25/2)(1^2)
100 = (225/2) - (25/2)
100 = 200/2 - 25/2
100 = 175/2
To simplify, we multiply both sides by 2:
200 = 175
This is a contradiction. Therefore, the point (3, 1) does not lie on the curve.
Since (3, 1) does not lie on the curve, we cannot find the equation of the tangent line at this point.