Question

How do you write a standard form equation for the hyperbola with $9{x^2} - 100{y^2} + 18x + 600y + 9 = 0$?

Answer 1

In order to determine the standard form equation of the hyperbola given the above question, first transpose the constant term toward RHS and then combine and group the terms by variable. Now try to complete the squares by applying some addition and subtraction for every group of variables. At the end divide both sides of the equation with the constant value on RHS to make the RHS equal to 1. You will get your required equation in form $\dfrac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} - \dfrac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} = 1$

Complete step by step answer:
We are given a equation of hyperbola as $9{x^2} - 100{y^2} + 18x + 600y + 9 = 0$.
Standard form of vertical hyperbola $\dfrac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} - \dfrac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} = 1$
In this question we have to write the given equation into the standard form of the hyperbola and to do so we will use completing the square method to combine term containing $x$into one single term and similarly with terms of$y$.
Let’s first transpose the constant term from left-hand side to right-hand side of the equation , equation becomes
$9{x^2} - 100{y^2} + 18x + 600y = - 9$
Grouping the like terms by variable and write them in the simplified form by pulling out common factors from them , we get
$\Rightarrow 9{x^2} + 18x - 100{y^2} + 600y = - 9 \\\ \Rightarrow 9\left( {{x^2} + 2x} \right) - 100\left( {{y^2} + 6y} \right) = - 9 \\\$
Now trying to complete the squares, by adding 9 and subtraction 900 from both sides of the equation ,
$\Rightarrow 9\left( {{x^2} + 2x} \right) + 9 - 100\left( {{y^2} + 6y} \right) - 900 = - 9 + 9 - 900$
Now again grouping the terms and simplifying further we get
$\Rightarrow 9\left( {{x^2} + 2x + 1} \right) - 100\left( {{y^2} + 6y + 9} \right) = - 900$
Now rewriting the above equation using the property ${A^2} + {B^2} + 2AB = {\left( {A + B} \right)^2}$by consider A as $x$and B as $1$in the first term of LHS and property ${A^2} + {B^2} - 2AB = {\left( {A - B} \right)^2}$by considering A as $y$and B as $3$for the second term in LHS. We get our equation as
$\Rightarrow 9{\left( {x + 1} \right)^2} - 100{\left( {y - 3} \right)^2} = - 900$
Now dividing both sides of the equation by $- 900$and simplifying it further , we get

$$\Rightarrow \dfrac{1}{{ - 900}}\left( {9{{\left( {x + 1} \right)}^2} - 100{{\left( {y - 3} \right)}^2}} \right) = \dfrac{1}{{ - 900}} \times - 900 \\\ \Rightarrow - \dfrac{{{{\left( {x + 1} \right)}^2}}}{{100}} + \dfrac{{{{\left( {y - 3} \right)}^2}}}{9} = 1 \\\ \Rightarrow \dfrac{{{{\left( {y - 3} \right)}^2}}}{9} - \dfrac{{{{\left( {x + 1} \right)}^2}}}{{100}} = 1 \\\$$

Hence we have obtained the equation of hyperbola in standard equation form as$\dfrac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} - \dfrac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} = 1$having centre at $\left( { - 1,3} \right)$
Therefore, the standard form equation of the given hyperbola is $\dfrac{{{{\left( {y - 3} \right)}^2}}}{9} - \dfrac{{{{\left( {x + 1} \right)}^2}}}{{100}} = 1$

Note: 1. When the centre of hyper is at the origin and foci are on the x-axis or y-axis , the Standard equation of hyperbola is
$\left[ {\left( {\dfrac{{{x^2}}}{{{a^2}}}} \right) - \left( {\dfrac{{{y^2}}}{{{b^2}}}} \right)} \right] = 1$
2.Make sure that the expansion of the terms is done carefully while determining the equation.
3.While completing the squares , be sure to keep both sides of the equation balanced.