Find an equation of the tangent line to the curve

2(x2+y2)2=25(x2−y2)

(a lemniscate) at the point (3,1). Please help.

take the derivative...

2(x^2+y^2+2xy)=25(x^2-y^2)
2(2x dx + 2ydy+2ydx+2xdy=25(2xdx-2ydy)

collect the dy terms

dy(4y+2x-50y)=dx(-4x-2y+50x)

dy/dx= you do it, then put in the point 3,1 and solve for dy/dx

To find the equation of the tangent line to the curve at the point (3,1), we first need to find the derivative of the curve.

The given curve is 2(x^2 + y^2)^2 = 25(x^2 - y^2).

Let's start by differentiating both sides of the equation with respect to x:

d/dx [2(x^2 + y^2)^2] = d/dx [25(x^2 - y^2)]

Using the chain rule on the left side, we have:

4(x^2 + y^2)(2x + 2y(dy/dx)) = 50x - 50y(dy/dx)

Next, we substitute the point (3, 1) into the equation to find the value of dy/dx at that point.

At (3, 1), we have:

4(3^2 + 1^2)(2(3) + 2(1)(dy/dx)) = 50(3) - 50(1)(dy/dx)

Simplifying this equation, we get:

4(10)(6 + 2(dy/dx)) = 50(3) - 50(1)(dy/dx)

40(6 + 2(dy/dx)) = 150 - 50(dy/dx)

240 + 80(dy/dx) = 150 - 50(dy/dx)

130(dy/dx) = 150 - 240

130(dy/dx) = -90

dy/dx = -90/130

dy/dx = -9/13

Now that we have the slope of the tangent line at the point (3, 1), we can use the point-slope form of a linear equation to find the equation of the tangent line.

The point-slope form is given by:

y - y1 = m(x - x1)

Substituting the values (3, 1) and dy/dx = -9/13 into the equation, we have:

y - 1 = (-9/13)(x - 3)

Simplifying this equation, we get:

y - 1 = (-9/13)x + (27/13)

y = (-9/13)x + (27/13) + 1

y = (-9/13)x + (27/13) + (13/13)

y = (-9/13)x + (40/13)

Therefore, the equation of the tangent line to the curve at the point (3, 1) is y = (-9/13)x + (40/13).

To find the equation of the tangent line to a curve at a given point, you need to follow these steps:

Step 1: Differentiate the equation of the curve implicitly.
Step 2: Plug in the coordinates of the given point into the derivative to find the slope of the tangent line.
Step 3: Use the point-slope form of a line to determine the equation of the tangent line.

Let's go through each step in detail.

Step 1: Differentiate the equation of the curve implicitly
Start by differentiating both sides of the equation with respect to x. This will give us the derivative of the equation.

Differentiating the left side:

d/dx[2(x^2 + y^2)^2] = d/dx[4(x^2 + y^2)(2x + 2y(dy/dx))]
= 4(x^2 + y^2)(2x + 2y(dy/dx)) + 4(x^2 + y^2)(2x + 2y(dy/dx))(dy/dx)

Differentiating the right side:

d/dx[25(x^2 - y^2)] = 50x - 50y(dy/dx)

Step 2: Plug in the coordinates of the given point into the derivative
We are given the point (3, 1), so substitute x = 3 and y = 1 into the derivative:

4(3^2 + 1^2)(2(3) + 2(1)(dy/dx)) + 4(3^2 + 1^2)(2(3) + 2(1)(dy/dx))(dy/dx) = 50(3) - 50(1)(dy/dx)

Expand and simplify the equation:

4(10)(6 + 2(dy/dx)) + 4(10)(6 + 2(dy/dx))(dy/dx) = 150 - 50(dy/dx)

Simplify further:

40(6 + 2(dy/dx)) + 40(6 + 2(dy/dx))(dy/dx) = 150 - 50(dy/dx)

Step 3: Use the point-slope form of a line to determine the equation of the tangent line
Now that we have the equation related to the slope of the tangent line, we can use the point-slope form of a line to find the equation of the tangent line at the point (3, 1).

The point-slope form is given by:

y - y1 = m(x - x1)

Where m is the slope of the tangent line, and (x1, y1) are the coordinates of the given point.

From the equation we obtained in Step 2, we can isolate the term dy/dx to find the slope:

40(6 + 2(dy/dx)) + 40(6 + 2(dy/dx))(dy/dx) + 50(dy/dx) = 150

Simplifying further:

240 + 80(dy/dx) + 80(dy/dx)(dy/dx) + 50(dy/dx) = 150

Rearranging the terms:

80(dy/dx)(dy/dx) + 130(dy/dx) + 90 = 0

Now, we have a quadratic equation in terms of dy/dx. We can solve this equation to find the value(s) of dy/dx.

Once you find the value(s) of dy/dx, substitute it back into the point-slope form equation, using the coordinates of the given point (3, 1), to find the equation of the tangent line.

With the equation of the tangent line, you can express it in either slope-intercept form (y = mx + b) or standard form (Ax + By = C).