determine the area under the hyperbola $y^2 −4x^2 = 1$

Question

Show that to find the length of an arc of the parabola $y = x^2$ one needs to determine the area under the hyperbola $y^2−4x^2 = 1$.

I solved for y and got $y=\sqrt {1+4x^2}$

But I really don't know what I am doing.

Answer 1

The starting point is that the arc lenght of a smoot curve of equation $y=f(x)$ from $x=a$ to $x=b$ is $$ s=\int_a^b\sqrt{1+(y'(x))^2} dx$$

so, for the arc lenght of the given parabola we have: $$ s=\int_a^b\sqrt{1+(2x)^2} dx$$

on the other side the positive branch of the given hyperbola is given by the equation: $$ y=\sqrt{1+4x^2} $$ so the area delimited by this curve and the $x-$ axis betveen $x=a$ and $x=b$ is: $$ A= \int_a^b\sqrt{1+4x^2}dx $$

and the two integrals a re the same.

Answer 2

The formula of arc length of the function $y=f(x)$ where $x\in[a,b]$ is $$ arc\ length=\int_{a}^{b}\sqrt{1+(f'(x))^2}\,dx $$ In your case $f(x)=x^2$, so $f'(x)=2x$. Therefore $$ arc\ length=\int_{a}^{b}\sqrt{1+(2x)^2}\,dx =\int_{a}^{b}\sqrt{1+4x^2}\,dx $$ which is the area under $y^2-4x^2=1$ in the interval $[a,b]$.