Question

The equation of a hyperbola is 5x2 − y2 = 25. What is the area of the asymptote rectangle?

Answer 1

`\bf \textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k) \end{cases}\\\\ -------------------------------\\\\ 5x^2-y^2=25\implies \cfrac{5x^2}{25}-\cfrac{y^2}{25}=1\implies \cfrac{x^2}{5}-\cfrac{y^2}{5^2}=1 \\\\\\ \cfrac{(x-0)^2}{(√(5))^2}-\cfrac{(y-0)^2}{5^2}=1\qquad \begin{cases} a=√(5)\\ b=5 \end{cases}`

now, the asymptotes rectangle is a rectangle that is (a+a) in width and (b+b) in length, thus its area is 2a*2b,


`\bf \stackrel{2a}{2√(5)}~~* ~~\stackrel{2b}{2\cdot 5}\implies 20√(5)`