Manipulate Ellipse Equations
Typically we are given the equation of an ellipse in the standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2+(y−k)2b2=1 or $\frac{\left(x-h\right)^2}{b^2}+\frac{\left(y-k\right)^2}{a^2}=1$(x−h)2b2+(y−k)2a2=1 where $a>b$a>b. Ideally we would like to express any equation of an ellipse in such a way, since we can immediately identify its centre and the length of the semi-major and semi-minor axes:
- $\left(h,k\right)$(h,k) is the centre of the ellipse.
- $a$a is the length of semi-major axis.
- $b$b is the length of semi-minor axis.
But generally speaking, the equation of an ellipse can be expressed in many ways. For instance, the following pair of equations both represent the same ellipse.
$9\left(x-2\right)^2+4\left(y-3\right)^2=36$9(x−2)2+4(y−3)2=36 or $9x^2-36x+4y^2-24x=-36$9x2−36x+4y2−24x=−36
In order to express the above equations in standard form, we can manipulate both sides by multiplying or by completing the square.
Worked examples
Example 1
Consider the equation of the ellipse $x^2-12x+36y^2=0$x2−12x+36y2=0
Rewrite the equation in the standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2+(y−k)2b2=1 where $a>b$a>b.
Think: Completing the square allows us to rewrite the quadratic expressions in $x$x as a perfect square of the form $\left(x-h\right)^2$(x−h)2. Notice that the quadratic expression in $y$y is already a perfect square.
To complete the square in $x$x we look at the expression $x^2-12x$x2−12x, halve and square the coefficient of $x$x, and then add the result to both sides of the equation.
Do: Halving and squaring $-12$−12 gives a result of $36$36, so we add this result to both sides of the equation:
$x^2-12x+36+36y^2=36$x2−12x+36+36y2=36
So altogether we get the following steps:
| $x^2-12x+36y^2$x2−12x+36y2 | $=$= | $0$0 | (Writing down the equation) |
| $x^2-12x+36+36y^2$x2−12x+36+36y2 | $=$= | $36$36 | (Completing the square for $x$x) |
| $\left(x-6\right)^2+36y^2$(x−6)2+36y2 | $=$= | $36$36 | (Factorising the perfect square) |
| $\frac{\left(x-6\right)^2}{36}+y^2$(x−6)236+y2 | $=$= | $1$1 | (Dividing both sides by $36$36) |
The equation of the ellipse is then $\frac{\left(x-6\right)^2}{36}+y^2=36$(x−6)236+y2=36. Optionally, the standard form can be made more obvious by writing it as $\frac{\left(x-6\right)^2}{6^2}+\frac{y^2}{1^2}=36$(x−6)262+y212=36.
Reflect: The equation of the ellipse centred at $\left(6,0\right)$(6,0) with a vertical semi-major axis of length $6$6 units and a horizontal semi-minor axis of length $1$1 unit can be expressed by the equation $x^2-12x+36y^2=0$x2−12x+36y2=0 or the equation $\frac{\left(x-6\right)^2}{36}+y^2=1$(x−6)236+y2=1.
Example 2
Consider the equation of the ellipse $9x^2-36x+4y^2-24x=-36$9x2−36x+4y2−24x=−36.
Rewrite the equation in the standard form $\frac{\left(x-h\right)^2}{b^2}+\frac{\left(y-k\right)^2}{a^2}=1$(x−h)2b2+(y−k)2a2=1 where $a>b$a>b.
Think: Completing the square allows us to rewrite the terms containing $x$x and $y$y in the form $\left(x-h\right)^2$(x−h)2 and $\left(y-k\right)^2$(y−k)2. Before we do so, it will be more convenient if we factorise so that the coefficients of $x^2$x2 and $y^2$y2 are $1$1.
Do: In this case, the coefficient of $x^2$x2 is $9$9 the coefficient of $y^2$y2 is $4$4:
$9\left(x^2-4x\right)+4\left(y^2-6x\right)=-36$9(x2−4x)+4(y2−6x)=−36
Now we can complete the square in $x$x and $y$y:
$9\left(x^2-4x+4\right)+4\left(y^2-6x+9\right)=-36+9\times4+4\times9$9(x2−4x+4)+4(y2−6x+9)=−36+9×4+4×9
Putting this together, we get the following steps:
| $9x^2-36x+4y^2-24x$9x2−36x+4y2−24x | $=$= | $-36$−36 | (Writing down the equation) |
| $9\left(x^2-4x\right)+4\left(y^2-6x\right)$9(x2−4x)+4(y2−6x) | $=$= | $-36$−36 | (Factorising the leading coefficients of each variable) |
| $9\left(x^2-4x+4\right)+4\left(y^2-6x+9\right)$9(x2−4x+4)+4(y2−6x+9) | $=$= | $-36+9\times4+4\times9$−36+9×4+4×9 | (Completing the square for $x$x and $y$y) |
| $9\left(x^2-4x+4\right)+4\left(y^2-6x+9\right)$9(x2−4x+4)+4(y2−6x+9) | $=$= | $36$36 | (Simplifying the constant terms) |
| $9\left(x-2\right)^2+4\left(y-3\right)^2$9(x−2)2+4(y−3)2 | $=$= | $36$36 | (Rewriting the perfect squares in factorised form) |
| $\frac{\left(x-2\right)^2}{4}+\frac{\left(y-3\right)^2}{9}$(x−2)24+(y−3)29 | $=$= | $1$1 | (Dividing both sides by $36$36) |
So the equation of the ellipse can be expressed in the form $\frac{\left(x-2\right)^2}{4}+\frac{\left(y-3\right)^2}{9}=1$(x−2)24+(y−3)29=1.
Reflect: The centre of the ellipse is $\left(h,k\right)=\left(2,3\right)$(h,k)=(2,3), the length of the vertical semi-major axis is $3$3 units and the length of the horizontal semi-minor axis is $2$2 units. From this, we can draw the equation of the ellipse as shown below.
 |
| Graph of the ellipse $9x^2-36x+4y^2-24x=-36$9x2−36x+4y2−24x=−36. |
Practice questions
question 1
Consider the ellipse with equation $\left(x+3\right)^2+4\left(y-1\right)^2=16$(x+3)2+4(y−1)2=16.
Write the equation in standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2+(y−k)2b2=1 where $a>b$a>b.
Draw the graph of the ellipse.
Loading Graph...
question 2
The graph of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$x29+y24=1 is given below.
Write $4x^2+9\left(y-2\right)^2=36$4x2+9(y−2)2=36 in the standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2+(y−k)2b2=1 where $a>b$a>b.
What transformation of the ellipse drawn gives the ellipse $4x^2+9\left(y-2\right)^2=36$4x2+9(y−2)2=36?
Translated to the right by $2$2 units
ATranslated to the left by $2$2 units
BTranslated up by $2$2 units
CTranslated down by $2$2 units
D
question 3
Consider the ellipse with equation $4x^2+24x+5y^2+20y+36=0$4x2+24x+5y2+20y+36=0.
Rewrite the equation in the standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2+(y−k)2b2=1 where $a>b$a>b.
What are the coordinates of the centre of the ellipse?
What are the coordinates of the vertices?