Show that the sum of the x- and y-intercepts of any tange...

[Mal]

Show that the sum of the x- and y-intercept of any tangent line to the curve sqrt[x]+sqrt[y]=sqrt[c] is equal to c.

I've figured that the x-intercept is where y=0, so sqrt[x]+sqrt[0]=sqrt[c], thus at the x-intercept y=c.
I've figured that the y-intercept is where x=0, so sqrt[0]+sqrt[y]=sqrt[c], thus at the y-intercept x=c.

I'm not sure if that also means x=y.

But I've also differentiated y with respect to x, and gotten y'=-[c/x^2].

I got that by dividing sqrt[x] from both sides in the original equation, getting sqrt[y]=sqrt[c/x], and then reasoning that y=c/x, and differentiating that.

Still, I'm not sure if what I've done so far is right, and I'm not sure where to go next even if it is.

 

[stapel]

Show that the sum of the x- and y-intercept of any tangent line to the curve sqrt[x]+sqrt[y]=sqrt[c] is equal to c.

I've figured that the x-intercept is where y=0, so sqrt[x]+sqrt[0]=sqrt[c], thus at the x-intercept y=c.

You need to find the intercepts of the tangent line, not the original curve.

I've also differentiated y with respect to x, and gotten y'=-[c/x^2].

I got that by dividing sqrt[x] from both sides in the original equation, getting sqrt[y]=sqrt[c/x], and then reasoning that y=c/x, and differentiating that.

Let's try differentiating the original curve instead, remembering that "c" is just some constant.

. . . . \(\displaystyle x^{\frac{1}{2}}\, +\, y^{\frac{1}{2}}\, =\, c^{\frac{1}{2}}\)

. . . . \(\displaystyle \frac{1}{2}x^{-\frac{1}{2}}\, +\, \frac{1}{2}y^{-\frac{1}{2}}\frac{dy}{dx}\, =\, 0\)

. . . . \(\displaystyle \frac{1}{\sqrt{x}}\, +\, \frac{1}{\sqrt{y}}\frac{dy}{dx}\, =\, 0\)

. . . . \(\displaystyle \frac{dy}{dx}\, =\, -\sqrt{\frac{y}{x}}\)

For any positive value x[sub:zopdjr3j]0[/sub:zopdjr3j], the corresponding y-value y[sub:zopdjr3j]0[/sub:zopdjr3j] (that is, the other coordinate of the point) on the original curve will be:

. . . . \(\displaystyle \sqrt{y_0}\, =\, \sqrt{c}\, -\, \sqrt{x_0}\)

. . . . \(\displaystyle y_0\, =\, c\, -\, 2\sqrt{cx_0}\, +\, x_0\)

Then the tangent line, for a given value of x[sub:zopdjr3j]0[/sub:zopdjr3j], will be of the form:

. . . . \(\displaystyle y\, -\, \left(c\, -\, 2\sqrt{cx_0}\, +\, x_0\right)\, =\, -\sqrt{\frac{y_0}{x_0}}\left(x\, -\, x_0\right)\)

What do you get if you plug "x = 0" and then "y = 0" into this equation? What is the sum?

 

[Mal]

Okay, I think I sort of get it. Using implicit differentiation, though, I believe that I was told that y was a function of x, so the real equation to differentiate would be sqrt[x] + sqrt[f(x)] = sqrt[c], right?

then I'd have to use the chain rule to differentiate sqrt[f(x)], and I'd get:

(1/2)x^(-1/2) + f'(x)*(1/2)(f(x))^(-1/2) = 0

And substituting y back in for f(x), that comes out as:

(1/2)x^(-1/2) + y' * (1/2)y^(-1/2) = 0

Is that the same things as the differentiation you offered? I assume y' = dy/dx, and I think your's said "(1/2)x^(-1/2) + (1/2)y^(-1/2) * dy/dx = 0"

I'm also a little confused about how you got to this:

For any positive value x[sub:b4wey5i7]0[/sub:b4wey5i7], the corresponding y-value y[sub:b4wey5i7]0[/sub:b4wey5i7] (that is, the other coordinate of the point) on the original curve will be:

. . . . \(\displaystyle \sqrt{y_0}\, =\, \sqrt{c}\, -\, \sqrt{x_0}\)

. . . . \(\displaystyle y_0\, =\, c\, -\, 2\sqrt{cx_0}\, +\, x_0\)

Where did these come from?

Also, what does y-sub 0 and x-sub 0 mean? It's not a notation I'm familiar with.