Question

How do you write the parabola $2{y^2} + 4y + x - 8 = 0$ in standard form and find the vertex, focus, and directrix?

Answer 1

Hint: In order to determine standard form equation of the parabola given the above question, first transpose all the terms toward RHS except terms containing variable $y$ and then combine and group the terms by variable. Now try to complete the squares by applying some addition and subtraction for the variable $y$. Now try to rewrite the result into the standard form and to focus , vertex and directrix use the standard results.

Complete step by step answer:
We are given an equation of Parabola as $2{y^2} + 4y + x - 8 = 0$.
Standard form of parabola which is quadratic in $y$ is of the form ${\left( {y - h} \right)^2} = 4a\left( {x - k} \right)$.
In this question we have to first write the given equation into the standard form of the parabola and to do so we will use completing the square method to combine terms containing $y$into one single term.
Let’s first transpose all the terms except term containing variable $y$from left-hand side to right-hand side of the equation , equation becomes
$2{y^2} + 4y = 8 - x$
Grouping the like terms and write them in the simplified form by pulling out common factors from them , we get
 \[ \Rightarrow 2\left( {{y^2} + 2y} \right) = 8 - x\]
Now trying to complete the squares, by adding 2 on both sides of the equation, we get
\[ \Rightarrow 2\left( {{y^2} + 2y} \right) + 2 = 8 - x + 2\]
Now again grouping the terms and simplifying further we get
\[ \Rightarrow 2\left( {{y^2} + 2y + 1} \right) = 10 - x\]
Now rewriting the above equation using the property ${A^2} + {B^2} + 2AB = {\left( {A + B} \right)^2}$by consider A as $y$and B as $1$in the term of LHS .We get our equation as
\[ \Rightarrow 2{\left( {y + 1} \right)^2} = - \left( {x - 10} \right)\]
Now dividing both sides of the equation by $2$and simplifying it further , we get
\[ \Rightarrow \dfrac{1}{2} \times 2{\left( {y + 1} \right)^2} = - \dfrac{1}{2} \times \left( {x - 10} \right)\]
Pulling out 4 as common from the Right-hand side of the equation, we have
\[ \Rightarrow {\left( {y - \left( { - 1} \right)} \right)^2} = 4\left( { - \dfrac{1}{8}} \right)\left( {x - 10} \right)\]
Hence we have obtained the equation in standard form.
Now comparing the above obtained result with the standard form of parabola ${\left( {y - h} \right)^2} = 4a\left( {x - k} \right)$, we get the values of variables as
\[
  h = 10 \\
  k = - 1 \\
  a = - \dfrac{1}{8} \\
 \]
As we know,
Centre$:\left( {h,k} \right)$$ = \left( {10, - 1} \right)$
Focus:
Focus for parabola is given as $\left( {h + a,k} \right)$
Putting the values we have
$
  \left( {10 + \left( { - \dfrac{1}{8}} \right), - 1} \right) \\
  \left( {10 - \dfrac{1}{8}, - 1} \right) \\
  \left( {\dfrac{{79}}{8}, - 1} \right) \\
 $
The Focus is $\left( {\dfrac{{79}}{8}, - 1} \right)$
 Directrix: It is given by $x = h - a = 10 - \left( { - \dfrac{1}{8}} \right) = \dfrac{{81}}{8}$
Therefore, the standard form equation of the given parabola is \[{\left( {y - \left( { - 1} \right)} \right)^2} = 4\left( { - \dfrac{1}{8}} \right)\left( {x - 10} \right)\]with
Centre$:\left( {h,k} \right)$$ = \left( {10, - 1} \right)$
Focus:$\left( {\dfrac{{79}}{8}, - 1} \right)$
Directrix: $x = \dfrac{{81}}{8}$

Note: 1. When the centre of parabola is at the origin, the Standard equation of parabola is
${\left( y \right)^2} = 4a\left( x \right)\,or\,{\left( x \right)^2} = 4a\left( y \right)$
2. Be careful about the signs. Use the values with proper sign as it may lead to wrong calculations
3.While completing the squares , be sure to keep both sides of the equation balanced.
4. Focus is a point from which all the points on the parabola are equal distance away.