# Math Help - Calculus problem with tangents

1. ## Calculus problem with tangents

I assume we have to use differentiation to answer this question, since it's what we're learning at the moment, the problem is, I have no idea where to begin with this question...

Consider the curve ${\sqrt{x} + \sqrt{y} = 1}$, for 0 (what "for 0" means, I have no clue).
Show that the sum of the x- and y- intercepts of the tangent line to any point (a,b) on the curve (a,b > 0), is equal to 1.

I expect a question like this may come up on an exam, so I'd really like to understand how to work it out. If I've posted this in the wrong section, I'm sincerely sorry. Any help will be very much appreciated.

2. Originally Posted by Exotique
I assume we have to use differentiation to answer this question, since it's what we're learning at the moment, the problem is, I have no idea where to begin with this question...

Consider the curve ${\sqrt{x} + \sqrt{y} = 1}$, for 0 (what "for 0" means, I have no clue).
"for x= 0" presumably.

Show that the sum of the x- and y- intercepts of the tangent line to any point (a,b) on the curve (a,b > 0), is equal to 1.

I expect a question like this may come up on an exam, so I'd really like to understand how to work it out. If I've posted this in the wrong section, I'm sincerely sorry. Any help will be very much appreciated.
If you are learning about the derivative, then you should have learned that the the derivative is the "slope of the tangent line".

To find the tangent line of this curve at x= a, first differentiate with respect to x. The equation is $x^{1/2}+ y^{1/2}= 1$ so differentiating with respect to x, $\frac{1}{2}x^{-1/2}+ \frac{1}{2}y^{-1/2}\frac{dy}{dx}= 0$ and solve for $\frac{dy}{dx}$. Evaluate it at x= a, y= b. (But note that a and b cannot be just any numbers. Since (a, b) is on that curve a and b must statisfy $\sqrt{a}+ \sqrt{b}= 1$.)

[tex]\frac{1}{2}x^{-1/2}+ \frac{1}{2}y^{-1/2}\frac{dy}{dx}= 0[/math leads to $y^{-1/2}\frac{dy}{dx}= -x^{-1/2}$ and then $\frac{dy}{dx}= -\frac{x^{-1/2}}{y^{-1/2}}= -\sqrt{\frac{y}{x}}$. At (a, b), that is $-\sqrt{\frac{b}{a}}$.

Of course, if the slope of the tangent line through (a, b) is m, the equation of the line is y= m(x- a)+ b so the equation of this line is $y= -\sqrt{\frac{b}{a}}(x- a)+ b$.

The x-intercept, the x value when y= 0, is given by $0= -\sqrt{\frac{b}{a}}(x- a)+ b$. Subtracting b from both sides, $-b= -\sqrt{\frac{b}{a}}(x- a)$. Dividing both sides by $-\sqrt{\frac{b}{a}}$, $b\sqrt{\frac{a}{b}}= \sqrt{ab}= x- a$ so x= $a+ \sqrt{ab}$.

The y-intercept, the y value when x= 0, is given by $y= -\sqrt{\frac{b}{a}}(0- a)+ b= \sqrt{ab}+ b$.

Now that gives sum $a+ b+ 2\sqrt{ab}$ but remember that $\sqrt{a}+ \sqrt{b}= 1$. What do you get if you square $\sqrt{a}+ \sqrt{b}$?

3. Thanks heaps for that! I had roughly that idea to begin with, but I didn't know how to put it down on paper.

Once again, thanks a lot!