How Do You Determine the Slope of a Tangent Line for the Curve 2cy = x^2 - c^2?

[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.

 

[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).

 

[EvLer]

What you want to do now is express c in terms of x and y (use the quadratic formula).

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

 

[AKG]

2cy = x^2 - c^2

This is a quadratic equation in c.

 

[EvLer]

2cy = x^2 - c^2

This is a quadratic equation in c.

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.

 

[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}

 

[EvLer]

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}

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

 

[AKG]

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

 

[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...

 

[HallsofIvy]

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}

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.

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

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

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

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

Thanks for any help.