# Curve Tracing I

• ### Introduction

Given an equation of a curve, say $y=f(x)$, the standard process of plotting involves obtaining many pairs of coordinates $(x,y)$, which satisfy the equation of curve. Having plotted a sufficiently large number of points, one gets a good picture of the curve.

In curve tracing (or curve sketching), we do not plot the points (generally) and get the exact shape of curve. However, using some mathematical facts and conclusions, we try to find an approximate shape of the curve. Thus, the figure so obtained may not always be to the scale and hence, the actual curve may look slightly different than the traced one. Nonetheless, we get an idea of how the curve looks like.

• ### Types of Equations

I) Equations in Cartesian Coordinates

${y^2 = 4ax, \ x= (y-1)(y-2)(y-3), \ x^2y + y^x = 6x}$

II) Equations in Polar Coordinates

${r = sin (a \theta), \ r^2 = a^2 cos (2 \theta), \ r = \frac {2a}{1+ cos (\theta)}}$

III) Parametric Equations

${x=at^2, y = 2at, \ x= acos^3 (\theta) , y= bsin^3 (\theta)}$

• ### General Observations (For equations in Cartesian Coordinates)

I) Points of Intersection with $X$ and $Y$ axes

By putting $x=0$, we get the point of intersection of the curve with $Y$-axis. (Similar procedure for $X$ axis and curve)

If there is no constant term in the equation of curve, it passes through the origin $(0,0)$.

II) Symmetry

III) Tangents

If the curve passes through the origin, the equation of tangent is obtaining by equating the lowest degree term to $0$; e.g. $y^2 =4ax$ passes through origin. The lowest degree term is $4ax$. Equating to $0$, $4ax =0$ or $x=0$, which is $Y$ axis.

To get the slope of tangent at a point $(x_1,y_1)$, we find $\frac {dy}{dx}$ at that point.

Recall : Parallel lines have same slope, slope of $X$ axis is $0$, slope of $Y$ axis is not defined. A positive (negative) value of slope implies an acute (obtuse) angle with $X$ axis.

IV) Asymptotes

An asymptote is a line which is tangential to the curve at infinity. e.g. $X$ and $Y$ axes are asymptotic to the rectangular hyperbola $xy = C$.

Asymptote parallel to $X$ ($Y$) axis is obtained by equating the highest degree term in $y$ ($x$) to $0$.

To get an oblique asymptote, we put $y=mx+c$ in the equation of the curve and get the values of $m$ and $c$.

• ### Illustrative Examples :

Q.1) Sketch the curve : ${y^2 = x^5 (2a-x)}$. (December 2015)

Remarks : (Assuming ${a>0}$)

1) Even powers of ${y}$ implies symmetry about ${X}$ axis.

2) Since ${LHS}$ is never negative, the region of absence of the curve will correspond to those values of ${x}$, for which ${RHS}$ is negative. When ${x<0}$, clearly ${x^5}$ will be negative and ${(2a-x)}$ will be positive. Hence to the left of the line ${x=0}$, the curve will not exist. (The Black Line)

3) Further, when ${x > 2a}$, ${x^5}$ will be positive and ${(2a-x)}$ will be negative. Hence to the right of the line ${x=2a}$, the curve will not exist. (The Blue Line)

4) The curve passes through the origin and intersects ${X}$ axis again at the point ${(2a,0)}$.

5) In the interval ${0 \le x \le 2a}$, values of ${y}$ are finite and hence there will be a peak (1st quadrant) and a trough (4th quadrant).

6) The tangent at origin will be obtained by equating the lowest degree term, i.e. ${y^2}$ to ${0}$. This gives 2 coincident lines ${y=0}$. Hence at ${(0,0)}$, the curve has a cusp.

7) At the point ${(2a,0)}$, the tangent will be parallel to ${Y}$ axis, since ${\frac {dy}{dx}}$ is not defined.

With all this information, the sketch of the curve will look like this :

(I’ve taken the value of ${a}$ as ${2}$).

Q.2)  Sketch the curve ${ay^2 = x^2(a-x)}$ (June 2015)

Remarks : (Assuming ${a > 0}$)

1) Even power of ${y}$ implies symmetry about ${X}$ axis.

2) Since ${LHS}$ is never negative, the region of absence of the curve will correspond to those values of ${x}$, for which ${RHS}$ is negative. For ${x > a}$, ${(a-x)}$ and hence ${RHS}$ will be negative. Hence, the curve will be absent to the right of the line ${x=a}$. (The Orange Line)

3) As ${x \to - \infty}$, ${x^2 (a-x)}$ will tend to ${\infty}$ and so will be ${y^2}$. Hence, in the second and third quadrants, the curve will extend to ${\infty}$.

4) The curve passes through origin. It intersects ${X}$ axis at the point ${(a,0)}$.

5) The lowest degree term is ${a (y^2-x^2)}$, so the equation(s) of tangent(s) at origin will be ${y-x=0}$ and ${y+x=0}$ i.e. ${y=x}$ and ${y=-x}$. (The Blue Line and The Green Line)

6) 2 tangents at origin imply that the curve passes through the origin twice.

With this information, the curve will look like this : (The Red Curve)

Q.3) Sketch the curve  ${y^2=x^2 (1-x)}$ (December 2013)

This is similar to Q.2, the only difference is the number ${1}$. In the previous problem, it was ${2}$.

Q.4) Sketch the curve ${x^2y^2 = a^2 (y^2-x^2)}$ (June 2014)

Remarks :

1) The curve passes through origin. It is symmetric about ${X}$ and ${Y}$ axes, because all powers are even.

2) ${LHS}$ is always non-negative. Hence in the region where ${y^2< x^2}$, i.e. ${y < x}$ or ${y > -x}$, the curve will be absent.

3) Equation of tangent to the curve at origin will be obtained by equating the lowest degree term to ${0}$, which in this case is ${y^2-x^2}$, which gives 2 tangents, ${y+x=0}$ and ${y-x = 0}$, i.e. ${y=- x}$ and ${y=x}$.

The line ${y=x}$ is the violet line and the line ${y=-x}$ is the black line.

4) From points 2 and 3, we can conclude that the region bounded by the lines which contains the ${X}$ axis is the region of absence of the curve, because in that region, ${y^2-x^2}$ is negative.

5) Rewriting the function, we get

${y^2 = \frac {-a^2 x^2}{x^2-a^2} = \frac {a^2 x^2}{a^2 - x^2}}$

Taking the limit as ${x \to a}$, we get the following :

${\lim \limits_{x \to a} \left | \frac {a^2x^2}{a^2-x^2} \right | = \infty}$

Hence the asymptotes parallel to ${Y}$ axis are ${x=a}$ and ${x=-a}$. (The Green Line and The Blue Line respectively)

Based on these points, the curve will look like this :

The value of ${a}$ taken by me is ${8}$.

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