# Slope of tangent line (DE)

1. Jun 27, 2005

### EvLer

Here's the problem:

detrmine DE giving the slope of the tangent line at point (x,y) for the given family of curves:

$$2cy = x^2 - c^2$$

Here's what I have:

slope of this curve:

$$y' = \frac{x}{c}$$

So, then the slope of tangent line is negative inverse:

$$y' = \m \frac{-c}{x}$$

so now I need c, but I don't see how to pull it out from the original equation!

Thanks for any help.

Last edited: Jun 27, 2005
2. Jun 27, 2005

### AKG

The negative reciprocal would be the slope of the perpendicular line. The slope of the tangent line would just be x/c. What you want to do now is express c in terms of x and y (use the quadratic formula).

3. Jun 27, 2005

### EvLer

That is precisely the problem.
and I'm not seeing how I can do that, even with quadratic formula.

4. Jun 27, 2005

### AKG

$$2cy = x^2 - c^2$$

This is a quadratic equation in c.

5. Jun 27, 2005

### EvLer

Are you talking about $$(a + b)^2$$? but there's y.
RHS has $$(a + b) (a - b)$$ but I don't know how that's helping me.

6. Jun 27, 2005

### AKG

A quadratic equation in x takes the form ax² + bx + c = 0, and the quadratic formula allows you to find the two complex solutions to this equation:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

7. Jun 27, 2005

### EvLer

Oh, yeah, complex... I didn't think of them being here.
Thanks a lot.

8. Jun 27, 2005

### AKG

Well your solutions for this quadratic equation will be real, but real numbers are complex numbers.

9. Jun 27, 2005

### EvLer

English terminology was the hardest part here: quadratic equation. It's called somth. else in my language and does't translate word for word, so I made a wrong association.
Thanks for your help, that is really like hischool algebra, I feel dumb....

10. Jun 27, 2005

### HallsofIvy

Staff Emeritus
You are completely off base here. You need to eliminate the constant c, getting a differential equation that does not involve c. Yes, 2cy'= 2x. Now, $$c^2+ 2cy= x^2$$ so $$c^2+ 2cy+ y^2= (c+y)^2= x^2+ y^2$$. That is: $$c+ y= \pm\sqrt{x^2+ y^2}$$
and $$c= -y+ \sqrt{x^2+ y^2}$$ or $$c= -y- \sqrt{x^2+ y^2}$$. The slope satisfies $$(-y+ \sqrt{x^2+ y^2})y'= x$$ or $$(-y- \sqrt{x^2+ y^2}= x$$.

No, the slope of the tangent line is the derivative. You are thinking of the slope of the normal line.

Last edited: Jun 27, 2005