# Thread: Finding the equation of a tangent line for an implicit function

1. ## Finding the equation of a tangent line for an implicit function

Given:

$\displaystyle xy^{2} + 5x^{2} = 4y$

Find the equation of the tangent line at x = 1

Say what?

Deriving the implicit function of y with respect to x yields:
$\displaystyle \frac{dy}{dx} = \frac{5}{y-2}$
but I'm completely stuck on how to find the equation of the tangent line from this (assuming I did this correctly).

This equation doesn't seem to be a function. I get multiple values of y for x equal 1.

Was I tricked?

2. Originally Posted by mbacarella
Given:

$\displaystyle xy^{2} + 5x^{2} = 4y$

Find the equation of the tangent line at x = 1

Say what?

Deriving the implicit function in terms of x yields
$\displaystyle \frac{dy}{dx} = \frac{5}{y-2}$
but I'm completely stuck on how to find the equation of the tangent line from this (assuming I did this correctly).
Find $\displaystyle f(1)$ (note this is the original function!) to give you the corresponding $\displaystyle y$ coordinate.

You can then sub the ordered pair $\displaystyle (1, f(1))$ into $\displaystyle \frac{dy}{dx}$

3. ## My algebra is broken

Thanks.

btw, that implicit function should have read:

$\displaystyle xy^{2} - 5x^{2} = 4y$

Anyway, here's what I got:
$\displaystyle (1, f(1)) = (1, (1)y^{2} - 5(1)^{2} = 4y) = (1, -1)$

Re-evaluating the derivative:
$\displaystyle \frac{d}{dx}( xy^{2} -5x^{2} = 4y ) = \frac{dx}{dx}2y^{1}\frac{dy}{dx} - 10x\frac{dx}{dx} =4y^{0}\frac{dy}{dx} = (1)2y\frac{dy}{dx} - 10x(1) = 4(1)\frac{dy}{dx}$

Now substituting y and x
$\displaystyle = (1)2(-1)\frac{dy}{dx} - 10(1) = 4(1)\frac{dy}{dx} = \frac{dy}{dx} = \frac{10}{2(-1)-4} = -\frac{5}{3}$

AKA
$\displaystyle f^{\prime}(1) = m_{tan} = -\frac{5}{3}$

Using the point-slope form
$\displaystyle y - y_{1} = m_{tan}(x - x_1) = y - (-1) = -\frac{5}{3}(x - 1) = y = -\frac{5}{3}x + \frac{5}{3} - 1$

Or finally?

$\displaystyle y = -\frac{5}{3}x + \frac{2}{3}$

I'd appreciate any comments or criticism on my thought processes or approach, even if the answer isn't wrong.

4. For

Use the product rule on the first term

y^2 - 2xydy/dx -10 x = 4dy/dx

dy/dx(-2xy-4) = 10x - y^2

dy/dx = [ y^2 -10 x]/[2xy+4]

When x =1 y^2 -4y -5 = 0 (y-5)(y +1) = 0

There are 2 possibilities (1,5) and (1,-1)

Now finish

5. ## Oops

Thanks for the correction. How did I miss that?

From the implicit derivation:
$\displaystyle \frac{d}{dx}\left[ xy^{2} -5x^{2} = 4y \right]$

$\displaystyle \frac{dx}{dx}y^{2} + x2y^{1}\frac{dy}{dx} - 10x\frac{dx}{dx} =4y^{0}\frac{dy}{dx}$

$\displaystyle (1)y^2 + 2xy\frac{dy}{dx} - 10x(1) = 4(1)\frac{dy}{dx}$

$\displaystyle y^2 + 2xy\frac{dy}{dx} - 10x = 4\frac{dy}{dx}$

$\displaystyle 2xy\frac{dy}{dx} - 4\frac{dy}{dx} = 10x - y^{2}$

$\displaystyle \frac{dy}{dx}(2xy - 4) = 10x-y^2$

We seem to disagree on this point:
$\displaystyle \frac{dy}{dx} = \frac{10x-y^{2}}{2xy-4}$

6. yes--- I had a sign mistake and have changed my last post to reflect this.

7. Ah, thank you.

Substituting for x and y:
$\displaystyle \frac{dy}{dx} = m_{tan} = \frac{10(1) - (-1)^{2}}{2(1)(-1)-4} = -\frac{9}{6} = -\frac{3}{2}$

Equation of a tangent line in point-slope form
$\displaystyle y - y_{1} = m_{tan}(x - x_1)$

Performing the substitutions:
$\displaystyle y - (-1) = -\frac{3}{2}(x - 1)$
$\displaystyle y = -\frac{3}{2}x + \frac{3}{2} - 1$

Or finally?
$\displaystyle y = -\frac{3}{2}x + \frac{1}{2}$

I'm curious why you said there were two possibilities (1, -1) and (1, 5). Wouldn't that make the dependent value y not a function?

8. ## n/m

For completeness sake:

I don't know why I expected to be unable to find tangent lines of relations. Please disregard my blathering on "not a function".