Find the slope of the tangent line to the curve (a lemniscate)

2(x^2+y^2)^2 = 25(x^2-y^2)
it is at pt (3,1)

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

:4[(3)² + (1)²][2(3) + 2(1)dy/dx] = 25[2(3) - 2(1)dy/dx]

:4[9 + (1)][6 + 2 dy/dx] = 25[6 - 2 dy/dx]

:4[10][8 dy/dx] = 25[4 dy/dx]

Divide both sides by 10

:4[8 dy/dx] = 10[dy/dx]
:32[dy/dx] = 10 [dy/dx]
then I got dy/dx = 10/32

You messed up near the end

look at line
4[9 + (1)][6 + 2 dy/dx] = 25[6 - 2 dy/dx]
and then
4[10][8 dy/dx] = 25[4 dy/dx]

you added "unlike terms" inside the bracket

should have been
40[6+2dy/dx] = 25[6-2dy/dx]
240 + 80dy/dx = 150 - 50dy/dx
130dy/dx = -90
dy/dx = -90/130 = -9/13

the rest is easy.

tyvm <3

To find the slope of the tangent line to the curve at a given point, you can use calculus.

Step 1: Start by finding the derivative of the given curve equation with respect to x.

Let's find the derivative of the curve equation:

2(x^2+y^2)^2 = 25(x^2-y^2)

We can rewrite the equation as:

2(x^4 + 2x^2y^2 + y^4) = 25x^2 - 25y^2

Taking the derivative of both sides with respect to x, we get:

8x^3 + 8xy^2(2y') + 4y^2(2x) = 50x - 50y^2(2y')

Simplifying it further, we have:

8x^3 + 16x^2y^2y' + 4y^2x = 50x - 100y^2y'

Step 2: Plug in the x-coordinate of the given point into the derivative equation.

We are given the point (3,1), so we substitute x = 3 into the equation.

8(3)^3 + 16(3)^2(1)^2y' + 4(1)^2(3) = 50(3) - 100(1)^2y'

Simplifying this equation further gives:

216 + 144y' + 12 = 150 - 100y'

Step 3: Solve the equation for y'.

Rearrange the equation to isolate the y' term:

144y' + 100y' = 150 - 216 + 12

244y' = -54

Divide both sides by 244:

y' = -54/244

Simplifying the fraction:

y' = -27/122

So, the slope of the tangent line to the curve (lemniscate) at the point (3,1) is -27/122.