finding the tangents and coordinates The point P, with coordinates $(p,q)$, lies on the graph of $x^\frac{1}{2} + y^\frac{1}{2} = a^\frac{1}{2}, a>0$. The tangent to the curve at $P$ cuts the axes at $(0,m)$ and $(n,0)$. Show that $m+n=a$  
I know the first step is finding the first derivative of the graph and equate to zero. but how do I find m and n?
 A: We need the slope of the tangent line at a generic point $(p,q)$ on the curve.  Differentiate implicitly. You should find that $\frac{dy}{dx}=-\frac{y^{1/2}}{x^{1/2}}$, so the derivative at $(p,q)$  is $-\frac{q^{1/2}}{p^{1/2}}$.  The equation of the tangent line at $(p,q)$ is therefore
$$y-q=-\frac{q^{1/2}}{p^{1/2}}(x-p),$$
which simplifies to 
$$yp^{1/2}+xq^{1/2}=qp^{1/2}+pq^{1/2}.$$  Note the symmetry. 
Now we find the intercepts. The $y$-intercept, which the problem calls $m$, is $q+p^{1/2}q^{1/2}$, and the $x$-intercept $n$ is
$q^{1/2}p^{1/2}+p$. 
Add. We get that the sum $m+n$ of the intercepts is $p+2p^{1/2}q^{1/2}+q$, which is $(p^{1/2}+q^{1/2})^2$.  But since $(p,q)$ is on the curve, we have $p^{1/2}+q^{1/2}=a^{1/2}$, and it's over.  
Remarks: $1.$ We could have first solved for $y$ in terms of $x$, and found the equation of the tangent line at the point with $x$-coordinate $p$. The calculations are in principle much like the ones above, but things definitely look messier. 
$2.$ We used the calculus. But if we want to avoid that, we can note that the line through $(0,m)$ and $(n,0)$ should meet the curve at a "double" point. Then algebra will give us the desired result.
