NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Manipulate Ellipse Equations
Lesson

Typically we are given the equation of an ellipse in the standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2+(yk)2b2=1 or $\frac{\left(x-h\right)^2}{b^2}+\frac{\left(y-k\right)^2}{a^2}=1$(xh)2b2+(yk)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(x2)2+4(y3)2=36 or $9x^2-36x+4y^2-24x=-36$9x236x+4y224x=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 $x^2-12x+36y^2=0$x212x+36y2=0

Rewrite the equation in the standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2+(yk)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 $\left(x-h\right)^2$(xh)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 $x^2-12x$x212x, 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:

$x^2-12x+36+36y^2=36$x212x+36+36y2=36

So altogether we get the following steps:

$x^2-12x+36y^2$x212x+36y2 $=$= $0$0 (Writing down the equation)
$x^2-12x+36+36y^2$x212x+36+36y2 $=$= $36$36 (Completing the square for $x$x)
$\left(x-6\right)^2+36y^2$(x6)2+36y2 $=$= $36$36 (Factorising the perfect square)
$\frac{\left(x-6\right)^2}{36}+y^2$(x6)236+y2 $=$= $1$1 (Dividing both sides by $36$36)

The equation of the ellipse is then $\frac{\left(x-6\right)^2}{36}+y^2=36$(x6)236+y2=36. Optionally, the standard form can be made more obvious by writing it as $\frac{\left(x-6\right)^2}{6^2}+\frac{y^2}{1^2}=36$(x6)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 $x^2-12x+36y^2=0$x212x+36y2=0 or the equation $\frac{\left(x-6\right)^2}{36}+y^2=1$(x6)236+y2=1.

Example 2

Consider the equation of the ellipse $9x^2-36x+4y^2-24x=-36$9x236x+4y224x=36.

Rewrite the equation in the standard form $\frac{\left(x-h\right)^2}{b^2}+\frac{\left(y-k\right)^2}{a^2}=1$(xh)2b2+(yk)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 $\left(x-h\right)^2$(xh)2 and $\left(y-k\right)^2$(yk)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(x24x)+4(y26x)=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(x24x+4)+4(y26x+9)=36+9×4+4×9

Putting this together, we get the following steps:

$9x^2-36x+4y^2-24x$9x236x+4y224x $=$= $-36$36 (Writing down the equation)
$9\left(x^2-4x\right)+4\left(y^2-6x\right)$9(x24x)+4(y26x) $=$= $-36$36 (Factorising the leading coefficients of each variable)
$9\left(x^2-4x+4\right)+4\left(y^2-6x+9\right)$9(x24x+4)+4(y26x+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(x24x+4)+4(y26x+9) $=$= $36$36 (Simplifying the constant terms)
$9\left(x-2\right)^2+4\left(y-3\right)^2$9(x2)2+4(y3)2 $=$= $36$36 (Rewriting the perfect squares in factorised form)
$\frac{\left(x-2\right)^2}{4}+\frac{\left(y-3\right)^2}{9}$(x2)24+(y3)29 $=$= $1$1 (Dividing both sides by $36$36)

So the equation of the ellipse can be expressed in the form $\frac{\left(x-2\right)^2}{4}+\frac{\left(y-3\right)^2}{9}=1$(x2)24+(y3)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$9x236x+4y224x=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(y1)2=16.

  1. Write the equation in standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2+(yk)2b2=1 where $a>b$a>b.

  2. 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.

Loading Graph...

  1. Write $4x^2+9\left(y-2\right)^2=36$4x2+9(y2)2=36 in the standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2+(yk)2b2=1 where $a>b$a>b.

  2. What transformation of the ellipse drawn gives the ellipse $4x^2+9\left(y-2\right)^2=36$4x2+9(y2)2=36?

    Translated to the right by $2$2 units

    A

    Translated to the left by $2$2 units

    B

    Translated up by $2$2 units

    C

    Translated down by $2$2 units

    D

    Translated to the right by $2$2 units

    A

    Translated to the left by $2$2 units

    B

    Translated up by $2$2 units

    C

    Translated 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.

  1. Rewrite the equation in the standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2+(yk)2b2=1 where $a>b$a>b.

  2. What are the coordinates of the centre of the ellipse?

  3. What are the coordinates of the vertices?

Outcomes

M8-1

Apply the geometry of conic sections

91573

Apply the geometry of conic sections in solving problems

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