Question

Trace the curve:
y2(x-a)=x2(2a-x)

Answer 1

The given curve can be written as

$$y^2={x^2(2a-x) \over (x-a)}$$

1.Symmetry

The curve is not symmetrical about the y-axis.

The curve is symmetrical about the x-axis.

$$(-y)^2={x^2(2a-x) \over (x-a)}\Longrightarrow (-y)^2=y^2$$

The curve is not symmetrical in opposite quadrants.

The curve is not symmetrical about the line y=x.

2. Origin.

The curve passes through the origin

$$x=0 \Longrightarrow y=0$$

3. Intersection with the coordinate axes.

$$y-intercept: x=0\Longrightarrow y=0, point(0, 0)$$ $$x-intercept: y=0\Longrightarrow 0={x^2(2a-x) \over (x-a)}$$ $$x=0\ or\ x=2a$$ $$point(0, 0), point(2a, 0)$$

4. First derivative

Take derivative with respect to x of both sides of the equation and use the Chain rule

$${d \over dx}(y^2(x-a))={d \over dx}(x^2(2a-x))$$

$$2y(x-a){dy \over dx}+y^2=2x(2a-x)-x^2$$

$${dy \over dx}={2x(2a-x)-x^2-y^2 \over 2y(2a-x)}$$

Thus the tangent at (2a, 0) is parallel to y-axis.

5. Tangents at the origin

The equations of the tangents to the curve at the origin is obtained by equating the lowest degree terms in x and y

in the given equation to zero

$$xy^2-ay^2=2ax^2-x^3$$

$$-ay^2=2ax^2$$

$$2x^2=-y^2$$

No tangent at the origin.

6. Asymptote(s)

$$y^2(x-a)=2ax^2-x^3$$

Asymptote parallel to y-axis

$$Coefficient\ of\ y^2\ is\ (x-a).$$

$$x-a=0\implies x=a$$

x=a is the a symptote parallel to y-axis.

There is no horizontal asymptote.

There is no slant (oblique) asymptote.

7. Regions where no part of the curve lies.

$$y^2={x^2(2a-x) \over (x-a)}$$

$$y=\pm \sqrt{x^2(2a-x) \over (x-a)}$$

$${x^2(2a-x) \over (x-a)}\geq0 \Longrightarrow a< x \leq 2a$$

Note that x=0 is the point on the curve but no branches passes through (0, 0), such a point is called as isolated point.