Manipulating hyperbola equations

Typically we are given the equation of a hyperbola 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 $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(y−k)2a2​−(x−h)2b2​=1. Ideally, we would like to express any equation of a hyperbola in this form since we can immediately identify its centre, orientation, and other features.

But generally speaking, the equation of a hyperbola can be expressed in many ways. For instance, the following pair of equations both represent the same hyperbola:

$4\left(x-2\right)^2-25\left(y-3\right)^2=100$4(x−2)2−25(y−3)2=100 and $x^2-4x-y^2+6y=105$x2−4x−y2+6y=105

In order to express the above equations in the standard form, we can manipulate both sides by multiplying or by completing the square.

 

Worked examples

Example 1

Consider the equation of the hyperbola $x^2-4x-y^2+6y=105$x2−4x−y2+6y=105.

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.

Think: Completing the square allows us to rewrite the quadratic expressions in $x$x and $y$y to be perfect squares of 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 the coefficient of $y^2$y2 is $1$1.

$x^2-4x-\left(y^2-6y\right)=105$x2−4x−(y2−6y)=105

Then to complete the square in $x$x we look at the expression $x^2-4x$x2−4x, halve and square the coefficient of $x$x, and then add the result to both sides of the equation.

Do: Halving and squaring $-4$−4 gives a result of $4$4, so we add this result to both sides of the equation:

$x^2-4x+4-\left(y^2-6y\right)=105+4$x2−4x+4−(y2−6y)=105+4

Similarly, to complete the square in $y$y we add $9$9 inside the brackets:

$x^2-4x+4-\left(y^2-6y+9\right)=105+4-9$x2−4x+4−(y2−6y+9)=105+4−9

Because of the negative sign outside of the brackets, we will subtract $9$9 from the other side to balance the equation.

Putting this together, we get the following steps:

$x^2-4x-y^2+6y$x2−4x−y2+6y	$=$=	$105$105	(Writing down the equation)
$x^2-4x-\left(y^2-6y\right)$x2−4x−(y2−6y)	$=$=	$105$105	(Factoring the terms with $y$y)
$x^2-4x+4-\left(y^2-6y+9\right)$x2−4x+4−(y2−6y+9)	$=$=	$105+4-9$105+4−9	(Completing the square for $x$x and $y$y)
$x^2-4x+4-\left(y^2-6y+9\right)$x2−4x+4−(y2−6y+9)	$=$=	$100$100	(Simplifying the constant terms)
$\left(x-2\right)^2-\left(y-3\right)^2$(x−2)2−(y−3)2	$=$=	$100$100	(Factoring the perfect squares)
$\frac{\left(x-2\right)^2}{100}-\frac{\left(y-3\right)^2}{100}$(x−2)2100​−(y−3)2100​	$=$=	$1$1	(Dividing both sides by $100$100)

The equation of the hyperbola is then $\frac{\left(x-2\right)^2}{100}-\frac{\left(y-3\right)^2}{100}=1$(x−2)2100​−(y−3)2100​=1 which is in the 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.

Reflect: The hyperbola with equation $x^2-4x-y^2+6y=105$x2−4x−y2+6y=105 has a centre of $\left(2,3\right)$(2,3) with vertices $\left(-8,3\right)$(−8,3) and $\left(12,3\right)$(12,3). What are the coordinates of the foci?

Example 2

Consider the equation of the circle $3y^2+12y-x^2-6x=3$3y2+12y−x2−6x=3.

Rewrite the equation in the standard form $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(h-k\right)^2}{b^2}=1$(y−k)2a2​−(h−k)2b2​=1.

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 the coefficients of $x^2$x2 and $y^2$y2 were $1$1.

Do: In this case, the coefficient of $x^2$x2 is $-1$−1 and the coefficient of $y^2$y2 is $3$3, so we factor out these coefficients:

$3\left(y^2+4y\right)-\left(x^2+6x\right)=3$3(y2+4y)−(x2+6x)=3

Now we can complete the square in $x$x and $y$y:

$3\left(y^2+4y+4\right)-\left(x^2+6x+9\right)=3+3\times4-9$3(y2+4y+4)−(x2+6x+9)=3+3×4−9

Notice that we've added by $3\times4$3×4 and subtracted by $9$9 on the other side to balance the equation. This is needed because the terms that we've added on the left hand side are being multiplied by $3$3 and $-1$−1 respectively.

Putting this together, we get the following steps:

$3y^2+12y-x^2-6x$3y2+12y−x2−6x	$=$=	$3$3	(Writing down the equation)
$3\left(y^2+4y\right)-\left(x^2+6x\right)$3(y2+4y)−(x2+6x)	$=$=	$3$3	(Factoring the terms with $x$x and $y$y)
$3\left(y^2+4y+4\right)-\left(x^2+6x+9\right)$3(y2+4y+4)−(x2+6x+9)	$=$=	$3+3\times4-9$3+3×4−9	(Completing the square for $x$x and $y$y)
$3\left(y^2+4y+4\right)-\left(x^2+6x+9\right)$3(y2+4y+4)−(x2+6x+9)	$=$=	$6$6	(Simplifying the constant terms)
$3\left(y+2\right)^2-\left(x+3\right)^2$3(y+2)2−(x+3)2	$=$=	$6$6	(Factoring the perfect squares)
$\frac{\left(y+2\right)^2}{2}-\frac{\left(x+3\right)^2}{6}$(y+2)22​−(x+3)26​	$=$=	$1$1	(Dividing both sides by $6$6)

So the equation of the hyperbola can be expressed in the form $\frac{\left(y+2\right)^2}{2}-\frac{\left(x+3\right)^2}{6}=1$(y+2)22​−(x+3)26​=1 which is in the form $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(h-k\right)^2}{b^2}=1$(y−k)2a2​−(h−k)2b2​=1.

Reflect: The centre of the hyperbola is $3y^2+12y-x^2-6x=3$3y2+12y−x2−6x=3 is $\left(-3,-2\right)$(−3,−2) and has vertices $\left(-3,-2+\sqrt{2}\right)$(−3,−2+√2) and $\left(-3,-2-\sqrt{2}\right)$(−3,−2−√2). What are the coordinates of the foci?

 

Practice questions

question 1

question 2

question 3