Question Statement

Find the focus, vertex, axis of symmetry, directrix, length of latus rectum, and end points of the latus rectum of the parabolas with the given equations. Also, draw the graph of each parabola.

(i) $y^2=6x$ (ii) $y^2=\frac{-3}{2}x$ (iii) $x^2=24y$ (iv) $x^2=-5y$ (v) $y^2-2y-12x-71=0$ (vi) $3x^2+42x+y+149=0$ (vii) $4y^2+4y=15-32x$ (viii) $9x^2-6x=108y+26$

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Background and Explanation

A parabola is the set of all points in a plane that are equidistant from a fixed point (the focus) and a fixed line (the directrix). The standard forms are $y^2=\pm 4ax$ (horizontal axis) and $x^2=\pm 4ay$ (vertical axis). For general quadratic equations, we use the method of "completing the square" to shift the equation into a standard form $(y-k{)}^2=4a(x-h)$ or $(x-h{)}^2=4a(y-k)$, where $(h,k)$ is the vertex.

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Solution

Part (i)

Equation: $y^2=6x$

Compare the given equation $y^2=6x$ with the standard form $y^2=4ax$: $\begin{array}{rl}& 4a=6\\ \Rightarrow & a=\frac{3}{2}\end{array}$

• Focus: The focus of this parabola is $(a,0)=\left(\frac{3}{2},0\right)$.
• Vertex: The vertex is at the origin $(0,0)$.
• Axis of symmetry: The axis of symmetry is the $x$-axis, and its equation is $y=0$.
• Directrix: The directrix is $x=-a\Rightarrow x=-\frac{3}{2}$ or $2x+3=0$.
• Length of latus rectum: $L=4a=4\left(\frac{3}{2}\right)=6$ units.
• End points of latus rectum: These are $(a,-2a)$ and $(a,2a)$, which are $\left(\frac{3}{2},-3\right)$ and $\left(\frac{3}{2},3\right)$.

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Part (ii)

Equation: $y^2=\frac{-3}{2}x$

Compare $y^2=\frac{-3}{2}x$ with $y^2=-4ax$: $\begin{array}{rl}& 4a=\frac{3}{2}\\ \Rightarrow & a=\frac{3}{8}\end{array}$

• Focus: The focus is $(-a,0)=\left(-\frac{3}{8},0\right)$.
• Vertex: The vertex is $(0,0)$.
• Axis of symmetry: The $x$-axis is the axis of symmetry ($y=0$).
• Directrix: The directrix is $x=a\Rightarrow x=\frac{3}{8}$ or $8x-3=0$.
• Length of latus rectum: $4a=4\left(\frac{3}{8}\right)=\frac{3}{2}$ units.
• End points of latus rectum: $(-a,-2a)$ and $(-a,2a)$, which are $\left(-\frac{3}{8},-\frac{3}{4}\right)$ and $\left(-\frac{3}{8},\frac{3}{4}\right)$.

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Part (iii)

Equation: $x^2=24y$

Compare $x^2=24y$ with $x^2=4ay$: $4a=24\Rightarrow a=6$

• Focus: The focus is $(0,a)=(0,6)$.
• Vertex: The vertex is $(0,0)$.
• Axis of symmetry: The $y$-axis is the axis of symmetry ($x=0$).
• Directrix: The directrix is $y=-a\Rightarrow y=-6$ or $y+6=0$.
• Length of latus rectum: $4a=4(6)=24$ units.
• End points of latus rectum: $(-2a,a)$ and $(2a,a)$, which are $(-12,6)$ and $(12,6)$.

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Part (iv)

Equation: $x^2=-5y$

Compare $x^2=-5y$ with $x^2=-4ay$: $4a=5\Rightarrow a=\frac{5}{4}$

• Focus: The focus is $(0,-a)=\left(0,-\frac{5}{4}\right)$.
• Vertex: The vertex is $(0,0)$.
• Axis of symmetry: The $y$-axis is the axis of symmetry ($x=0$).
• Directrix: The directrix is $y=a\Rightarrow y=\frac{5}{4}$ or $4y-5=0$.
• Length of latus rectum: $4a=4\left(\frac{5}{4}\right)=5$ units.
• End points of latus rectum: $(-2a,-a)$ and $(2a,-a)$, which are $\left(-\frac{5}{2},-\frac{5}{4}\right)$ and $\left(\frac{5}{2},-\frac{5}{4}\right)$.

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Part (v)

Equation: $y^2-2y-12x-71=0$

First, complete the square for $y$: $\begin{array}{rl}{y}^2-2y& =12x+71\\ y^2-2y+1& =12x+71+1\\ (y-1{)}^2& =12(x+6)\end{array}$ Let $Y=y-1$ and $X=x+6$. The equation becomes $Y^2=12X$. Comparing with $Y^2=4aX$, we find $4a=12\Rightarrow a=3$.

• Vertex: Set $X=0$ and $Y=0$. $x+6=0\Rightarrow x=-6$ and $y-1=0\Rightarrow y=1$. Vertex is $(-6,1)$.
• Focus: $(X,Y)=(a,0)\Rightarrow x+6=3\Rightarrow x=-3$ and $y-1=0\Rightarrow y=1$. Focus is $(-3,1)$.
• Axis of symmetry: $Y=0\Rightarrow y-1=0$.
• Directrix: $X=-a\Rightarrow x+6=-3\Rightarrow x+9=0$.
• Length of latus rectum: $4a=12$ units.
• End points of latus rectum: $(X,Y)=(a,\pm 2a)$.
  • For $(3,6)$: $x+6=3\Rightarrow x=-3$ and $y-1=6\Rightarrow y=7$.
  • For $(3,-6)$: $x+6=3\Rightarrow x=-3$ and $y-1=-6\Rightarrow y=-5$.
  • Points are $(-3,7)$ and $(-3,-5)$.

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Part (vi)

Equation: $3x^2+42x+y+149=0$

Complete the square for $x$: $\begin{array}{rl}3(x^2+14x)& =-y-149\\ 3(x^2+14x+49)& =-y-149+3(49)\\ 3(x+7{)}^2& =-y-149+147\\ 3(x+7{)}^2& =-(y+2)\\ (x+7{)}^2& =-\frac{1}{3}(y+2)\end{array}$ Let $X=x+7$ and $Y=y+2$. Equation is $X^2=-\frac{1}{3}Y$. Comparing with $X^2=-4aY$, we find $4a=\frac{1}{3}\Rightarrow a=\frac{1}{12}$.

• Vertex: $X=0,Y=0\Rightarrow (-7,-2)$.
• Focus: $(X,Y)=(0,-a)\Rightarrow x+7=0\Rightarrow x=-7$ and $y+2=-\frac{1}{12}\Rightarrow y=-\frac{25}{12}$. Focus is $\left(-7,-\frac{25}{12}\right)$.
• Axis of symmetry: $X=0\Rightarrow x+7=0$.
• Directrix: $Y=a\Rightarrow y+2=\frac{1}{12}\Rightarrow 12y+23=0$.
• Length of latus rectum: $4a=\frac{1}{3}$ units.
• End points of latus rectum: $(X,Y)=(\pm 2a,-a)$.
  • $X=\pm 2(\frac{1}{12})=\pm \frac{1}{6}$.
  • $x+7=\frac{1}{6}\Rightarrow x=-\frac{41}{6}$; $x+7=-\frac{1}{6}\Rightarrow x=-\frac{43}{6}$.
  • $y=-\frac{25}{12}$.
  • Points are $\left(-\frac{41}{6},-\frac{25}{12}\right)$ and $\left(-\frac{43}{6},-\frac{25}{12}\right)$.

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Part (vii)

Equation: $4y^2+4y=15-32x$

Complete the square for $y$: $\begin{array}{rl}4(y^2+y+\frac{1}{4})& =15-32x+1\\ 4(y+\frac{1}{2}{)}^2& =16-32x\\ (y+\frac{1}{2}{)}^2& =4-8x=-8(x-\frac{1}{2})\end{array}$ Let $Y=y+\frac{1}{2}$ and $X=x-\frac{1}{2}$. Equation is $Y^2=-8X$. Comparing with $Y^2=-4aX$, we find $4a=8\Rightarrow a=2$.

• Vertex: $X=0,Y=0\Rightarrow (\frac{1}{2},-\frac{1}{2})$.
• Focus: $(X,Y)=(-a,0)\Rightarrow x-\frac{1}{2}=-2\Rightarrow x=-\frac{3}{2}$ and $y=-\frac{1}{2}$. Focus is $\left(-\frac{3}{2},-\frac{1}{2}\right)$.
• Axis of symmetry: $Y=0\Rightarrow 2y+1=0$.
• Directrix: $X=a\Rightarrow x-\frac{1}{2}=2\Rightarrow 2x-5=0$.
• Length of latus rectum: $4a=8$ units.
• End points of latus rectum: $(X,Y)=(-a,\pm 2a)=(-2,\pm 4)$.
  • $x-\frac{1}{2}=-2\Rightarrow x=-\frac{3}{2}$.
  • $y+\frac{1}{2}=4\Rightarrow y=\frac{7}{2}$; $y+\frac{1}{2}=-4\Rightarrow y=-\frac{9}{2}$.
  • Points are $\left(-\frac{3}{2},\frac{7}{2}\right)$ and $\left(-\frac{3}{2},-\frac{9}{2}\right)$.

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Part (viii)

Equation: $9x^2-6x=108y+26$

Complete the square for $x$: $\begin{array}{rl}9(x^2-\frac{2}{3}x+\frac{1}{9})& =108y+26+1\\ 9(x-\frac{1}{3}{)}^2& =108y+27\\ 9(x-\frac{1}{3}{)}^2& =108(y+\frac{1}{4})\\ (x-\frac{1}{3}{)}^2& =12(y+\frac{1}{4})\end{array}$ Let $X=x-\frac{1}{3}$ and $Y=y+\frac{1}{4}$. Equation is $X^2=12Y$. Comparing with $X^2=4aY$, we find $4a=12\Rightarrow a=3$.

• Vertex: $X=0,Y=0\Rightarrow (\frac{1}{3},-\frac{1}{4})$.
• Focus: $(X,Y)=(0,a)\Rightarrow x=\frac{1}{3}$ and $y+\frac{1}{4}=3\Rightarrow y=\frac{11}{4}$. Focus is $\left(\frac{1}{3},\frac{11}{4}\right)$.
• Axis of symmetry: $X=0\Rightarrow 3x-1=0$.
• Directrix: $Y=-a\Rightarrow y+\frac{1}{4}=-3\Rightarrow 4y+13=0$.
• Length of latus rectum: $4a=12$ units.
• End points of latus rectum: $(X,Y)=(\pm 2a,a)=(\pm 6,3)$.
  • $x-\frac{1}{3}=6\Rightarrow x=\frac{19}{3}$; $x-\frac{1}{3}=-6\Rightarrow x=-\frac{17}{3}$.
  • $y=\frac{11}{4}$.
  • Points are $\left(\frac{19}{3},\frac{11}{4}\right)$ and $\left(-\frac{17}{3},\frac{11}{4}\right)$.

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Key Formulas or Methods Used

• Standard Forms:
  • $y^2=4ax$: Focus $(a,0)$, Directrix $x=-a$.
  • $y^2=-4ax$: Focus $(-a,0)$, Directrix $x=a$.
  • $x^2=4ay$: Focus $(0,a)$, Directrix $y=-a$.
  • $x^2=-4ay$: Focus $(0,-a)$, Directrix $y=a$.
• Completing the Square: $a{x}^2+bx+c=a(x+\frac{b}{2a}{)}^2+(c-\frac{b^2}{4a})$.
• Latus Rectum: Length is always $|4a|$. Endpoints are found by substituting the focus coordinate into the equation.

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Summary of Steps

1. Rearrange the equation: Group the squared variable and its linear term on one side.
2. Complete the square: Add and subtract the necessary constant to form a perfect square trinomial.
3. Identify the standard form: Determine if the parabola opens up, down, left, or right.
4. Find the value of $a$: Equate the coefficient of the linear side to $4a$.
5. Calculate properties: Use $a$ and the vertex $(h,k)$ to find the focus, directrix, axis, and latus rectum.
6. Graph: Plot the vertex, focus, and latus rectum endpoints to sketch the curve.