Question

Trace the curve
a2y2=x2(a-x)

Answer 1

Let $a>0$ . The curve $G=\left\{ \left( x,y \right) \ :\ { a }^{ 2 }{ y }^{ 2 }{ =x }^{ 2 }{ (a-x) }^{ } \right\}$  is the union of the function of the graph $y(x)=\frac { x }{ a } \sqrt { a-x }$ and $y(x)=-\frac { x }{ a } \sqrt { a-x }$ defined on the set $(-\infty ,a]$ and symmetric about the $x-$ axis. Therefore we draw a curve according to the properties of the function $y(x)=\frac { x }{ a } \sqrt { a-x }$

$1)\quad ay'(x)=\sqrt { a-x } -\frac { x }{ 2\sqrt { a-x } } =\frac { 2a-3x }{ 2\sqrt { a-x } } \ ,x\ <a\quad.$

If $x<\frac { 2a }{ 3 }$, then $y'(x)>0$ and $y(x)\nearrow \quad$.

If $\frac { 2a }{ 3 } <x<a$ , then $y'(x)<0$ and $y(x)\searrow$

2) $2ay"(x)=\ \frac { d }{ dx } \left( \frac { 2a-3x }{ \ \sqrt { a-x } } \right)$$=\frac { -3\sqrt { a-x } +\ \frac { (2a-3x) }{ 2\sqrt { a-x } } \ \ \ }{ (a-x) } =$ $\frac { -6a+\ 6x+2a-3x }{ 2(a-x)\sqrt { a-x } \ } \ =\frac { -4a+3x }{ 2(a-x)\sqrt { a-x } }$

Since $x<a$ , then $-4a+3x<-4a+3a<0$ and $y"(x)<0$ .So function $y(x)$ is convex up.

3) The function graph has no vertical asymptotes , since the function is continuous. There no oblique asymptotes since the $\lim _{ x\rightarrow -\infty \ }{ \frac { f(x) }{ x } } =\lim _{ x\rightarrow -\infty \ }{ \frac { x\sqrt { a-x } \quad }{ ax } = } +\infty$