# Calculus help please?

Find the equation of the tangent line to the curve (a lemniscate) 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 what is the value of m? what is the value of b?

### 3 Answers

- Wayne DeguManLv 71 month ago
2(x² + y²)² = 25(x² - y²)

Differentiating w.r.t. x we get:

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

=> 4(x² + y²)(x.dx + y.dy) = 25(x.dx - y.dy)

Using x = 3 and y = 1 we get:

4(3² + 1²)(3.dx + dy) = 25(3.dx - dy)

=> 40(3.dx + dy) = 25(3.dx - dy)

i.e. 8(3.dx + dy) = 5(3.dx - dy)

=> 24.dx + 8.dy = 15.dx - 5.dy

so, 13.dy = -9.dx

Hence, dy/dx = -9/13

So, tangent equation is y = -9x/13 + b

Again, using x = 3 and y = 1 we get:

1 = -27/13 + b

so, b = 40/13

Hence, y = -9x/13 + 40/13

or, 13y = 40 - 9x

A sketch is shown below.

:)>

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- 1 month ago
2 * (x^2 + y^2)^2 = 25 * (x^2 - y^2)

Derive implicitly

2 * 2 * (x^2 + y^2) * (2x * dx + 2y * dy) = 25 * (2x * dx - 2y * dy)

Plug in values for x and y, then solve for dy/dx

x = 3 , y = 1

4 * (3^2 + 1^2) * (2 * 3 * dx + 2 * 1 * dy) = 25 * (2 * 3 * dx - 2 * 1 * dy)

4 * (9 + 1) * (6 * dx + 2 * dy) = 25 * (6 * dx - 2 * dy)

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

4 * (6 * dx + dy) = 5 * (3 * dx - dy)

24 * dx + 4 * dy = 15 * dx - 5 * dy

24 * dx - 15 * dx = -5 * dy - 4 * dy

9 * dx = -9 * dy

dx = -dy

1 = -dy/dx

-1 = dy/dx

dy/dx = m

(3 , 1) = (h , k)

y - k = m * (x - h)

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

y - 1 = -x + 3

y = -x + 4

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- JohnLv 71 month ago
I'd help but I haven't taken that since 1973, but I still remember the principle of limits.

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