Question

How do you put $ {x^2} - 2x + 2{y^2} - 12y + 3 = 0 $ in standard form, find the centre, the end-points, vertices, the foci and eccentricity?

Answer 1

Hint: In order to determine the standard form equation of the ellipse 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. Compare the result obtained with standard form $ \dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 $ to get the values for the variables.For Foci , centre,end-points,and eccentricity use the standard formulas.

Complete step-by-step answer:
We are given an equation of Ellipse as $ {x^2} - 2x + 2{y^2} - 12y + 3 = 0 $ .
Standard form of ellipse is of the form $ \dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 $
In this question we have to first write the given equation into the standard form of the ellipse 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
 $ {x^2} - 2x + 2{y^2} - 12y = - 3 $
Grouping the like terms by variable and write them in the simplified form by pulling out common factors from them , we get
 \[ \Rightarrow {x^2} - 2x + 2\left( {{y^2} - 6y} \right) = - 3\]
Now trying to complete the squares, by adding 1 and 18 on both sides of the equation ,
\[ \Rightarrow {x^2} - 2x + 1 + 2\left( {{y^2} - 6y} \right) + 18 = - 3 + 1 + 18\]
Now again grouping the terms and simplifying further we get
\[ \Rightarrow \left( {{x^2} - 2x + 1} \right) + 2\left( {{y^2} - 6y + 9} \right) = - 3 + 1 + 18\]
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 similarly using property by considering A as $ y $ and B as $ 3 $ for the second term in LHS. We get our equation as
\[ \Rightarrow {\left( {x - 1} \right)^2} + 2{\left( {y - 3} \right)^2} = 16\]
Now dividing both sides of the equation by $ 16 $ and simplifying it further , we get
\[
   \Rightarrow \dfrac{1}{{16}}\left( {{{\left( {x - 1} \right)}^2} + 2{{\left( {y - 3} \right)}^2}} \right) = \dfrac{1}{{16}} \times 16 \\
   \Rightarrow \dfrac{{{{\left( {x - 1} \right)}^2}}}{{16}} + \dfrac{{{{\left( {y - 3} \right)}^2}}}{8} = 1 \\
   \Rightarrow \dfrac{{{{\left( {x - 1} \right)}^2}}}{{{{\left( 4 \right)}^2}}} + \dfrac{{{{\left( {y - 3} \right)}^2}}}{{{{\left( {2\sqrt 2 } \right)}^2}}} = 1 \;
 \]
Centre $ :\left( {h,k} \right) $ $ = \left( {1,3} \right) $
Hence we have obtained the equation of Ellipse in standard equation form as $ \dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 $ .
Now,
Major axis is $ a = 4 $
Minor axis is $ b = 2\sqrt 2 $
End-Points: $ \left( {h \pm a,k} \right) $ , $ \left( {h,k \pm b} \right) $
 $ \left( {5,3} \right),\left( {1,3 + 2\sqrt 2 } \right),\left( { - 3,3} \right),\left( {1,3 - 2\sqrt 2 } \right) $
Eccentricity:
 $ e = \sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} $ ,putting values for $ {a^2} = 16,{b^2} = 8 $
 $
  e = \sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} \\
  e = \sqrt {\dfrac{{16 - 8}}{{16}}} = \sqrt {\dfrac{8}{{16}}} \\
  e = 0.707 \;
  $
Foci:
Calculating $ c $ from formula $ {c^2} = {a^2} - {b^2} $
  $
  {c^2} = 16 - 8 \\
  {c^2} = 8 \\
  c = 2\sqrt 2 \;
  $
So the foci for the ellipse will be $ \left( {h \pm c,k} \right) $
Foci\[ = \left( {1 + 2\sqrt 2 ,3} \right),\left( {1 - 2\sqrt 2 ,3} \right)\]
Therefore, the standard form equation of the given ellipse is \[\dfrac{{{{\left( {x - 1} \right)}^2}}}{{{{\left( 4 \right)}^2}}} + \dfrac{{{{\left( {y - 3} \right)}^2}}}{{{{\left( {2\sqrt 2 } \right)}^2}}} = 1\]with
Centre $ :\left( {h,k} \right) $ $ = \left( {1,3} \right) $
End-Points: $ \left( {5,3} \right),\left( {1,3 + 2\sqrt 2 } \right),\left( { - 3,3} \right),\left( {1,3 - 2\sqrt 2 } \right) $
Eccentricity:
 $ e = 0.707 $
Foci:\[\left( {1 + 2\sqrt 2 ,3} \right),\left( {1 - 2\sqrt 2 ,3} \right)\]

Note: 1. When the centre of ellipse is at the origin and foci are on the x-axis or y-axis , the Standard equation of ellipse 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.