Conic Sections (Parabola) - Part 3

Translated Parabola

Parabola တစ်ခု၏
  • vertex က $(0,0)$၊ focus က (0, p) ၌ ရှိသော parabola ကို $x^2=4py$ ဟူ၍ လည်းကောင်း

  • vertex က $(0,0)$၊ focus က (p, 0) ၌ ရှိသော parabola ကို $y^2=4px$ ဟူ၍ လည်းကောင်း
ဖော်ပြနိုင်ကြောင်း Conic Sections Part(1) နှင့် Part (2) တွင် တင်ပြခဲ့ပြီး ဖြစ်ပါသည်။

တဖန် Parabola တစ်ခု၏
  • standard form ကို $y=ax^2+bx+c$ ဟူ၍ လည်းကောင်း

  • vertex form with vertex at $(h,k)$ ကို $y=a(x-h)^2+k\Rightarrow y-k=a(x-h)^2$ ဟူ၍ လည်းကောင်း
ဖော်ပြနိုင်ကြောင်း grade (10) သင်ခန်းစားတွင် သိရှိခဲ့ပြီး ဖြစ်သည်။

ယခု vertex က $(h,k)$ ၌ ရှိသော translated parabola များအကြောင်း ဆက်လက် လေ့လာပါမည်။ vertex က $(h,k)$ ၌ ရှိသောအခါ

  • မူလ equation တွင်ရှိသော $x$ သည် $(x-h)$ သို့လည်းကောင်း

  • မူလ equation တွင်ရှိသော $y$ သည် $(y-k)$ သို့လည်းကောင်း

  • horizontal parabola $y^2=4px$ ကို $(y-k)^2=4p(x-h)$

  • vertical parabola $x^2=4py$ ကို $(x-h)^2=4p(y-k)$
ဟုပြောင်းပေးရမည်။

Vertex parabola $(h,k)$ ၌ ရှိသော translated parabola များ၏ general equation ကို အောက်ပါအတိုင်း မှတ်သားနိုင်ပါသည်။

Standard Forms of Equations of Parabolas with Vertex at $(h, k)$

Equation Vertex Axis of
Symmetry
Focus Directrix Endpoints of
Latus Rectum
Opens

$(y-k)^2=4p(x-h)$

$(h,k)$

Horizontal: $y=k$

$(h+p,k)$

$x=h-p$

$(h+p, k\pm 2p)$
RIGHT, $p > 0$

LEFT, $p < 0$

$(x-h)^{2}=4 p(y-k)$

$(h, k)$

Vertical: $x=h$

$(h,k+p)$

$y=k-p$

$(h\pm 2p, k+p)$
UP, $p > 0$

DOWN, $p < 0$

Note: The focus is distance $|p|$ from the vertex.




Graphs of parabolas with vertex at $(h, k)$ and $p > 0$


Worked Examples

Example (1)

Find the vertex, focus, and directrix of the parabola given by $(x - 3)^2 = 8(y + 1)$. Hence graph the parabola.

Solution

$(x - 3)^2 = 8(y + 1)$

Comparing with $(x - h)^2 = 4p(y - k)$, whe have

$h=3, k=-1$ and $4p = 8\Rightarrow p=2$

$\therefore\ \ \text{vertex} = (h, k) = (3, -1)$.

$\ \ \ \ \ \ \text{focus} = (h, k+p) = (3, 1)$.

$\ \ \ \ \ \ \text{directrix} : y= k-p = -1-2=-3$.

$\ \ \ \ \ \ \text{Endpoints of Latus rectum} = (h\pm 2p, k+p)$.

$\therefore\ \ \text{Endpoints of Latus rectum} = (-1,1)\ \text{and}\ (7,1)$.



Example (2)

Find the vertex, focus, and directrix of the parabola given by $y^2 + 2y + 12x - 23 = 0$. Hence graph the parabola.

Solution

$\begin{array}{l} {{y}^{2}}+2y+12x-23=0\\\\ {{y}^{2}}+2y=-12x+23\\\\ {{y}^{2}}+2y+1=-12x+24\\\\ {{(y+1)}^{2}}=-12(x-2)\\\\ \text{Comparing with}\ {{(y-k)}^{2}}=4p(x-h)\\\\ h=2,\ k=-1\\\\ \therefore \ \ \ \ \text{Vertex}\ \text{= }(2,\ -1)\\\\ \ \ \ \ \ \ 4p=-12\Rightarrow p=-3\\\\\therefore \ \ \ \ \text{Focus}\ \text{= }(h+p,\ k)=(-1,\ -1)\\\\ \ \ \ \ \ \ \text{Directrix}\ \text{:}\ x=h-p\Rightarrow x=5\\\\ \ \ \ \ \ \text{Endpoints of latus rectum}\ =(h+p,\ k\ \pm \ 2p)\\\\ \therefore \ \ \text{Endpoints of latus rectum}\ =(-1,\ -7)\ \text{and}\ (-1,\ 5) \end{array}$



Example (3)

Find the vertex and equation of parabola with its focus at $(3,2)$ and the directrix is $x=-1$. Hence sketch the graph. Solution

Focus $= (3, 2)$

Directrix : $x = -1$

Since the directrix is a vertical line it is horizontal parabola which opens to the right.

For a horizontal parabola, focus = $(h+p, k)$

$\therefore \ \ (h+p, k)=(3, 2)\Rightarrow h + p = 3\ \text{and}\ k=2$

For a horizontal parabola the equation of directrix is $x = h - p$.

$\therefore \ \ h - p = -1$

Solving the equations $h + p = 3$ and $h - p=-1$, we get $h=1$ and $p=2$.

$\therefore \ \ \text{vertex} = (1, 2)$.

$ \ \ \ \ \ \ \text{equation of parabola}: (y-k)^2 = 4p(x-h)\Rightarrow (y-2)^2=8(x-1)^2$

To sketch the graph, we need to determine the endpoints of latus rectum.

Endpoints of latus rectum = $(h + p, k\pm 2p)$

Therefore the endpoints of latus rectum are $(3,-2 )$ and $(3, 6)$.



Exercises

  1. Find the focus and directrix of the parabola with the given equation. Hence graph the parabola.

    $\begin {array}{ll}
    \text{(a)}& y^{2}=16 x\\
    \text{(b)}& y^{2}=4 x\\
    \text{(c)}& y^{2}=-8 x\\
    \text{(d)}& y^{2}=-12 x\\
    \text{(e)}& x^{2}=12 y\\
    \text{(f)}& x^{2}=8 y\\
    \text{(g)}& x^{2}=-16 y\\
    \text{(h)}& x^{2}=-20 y\\
    \text{(i)}& y^{2}-6 x=0\\
    \text{(j)}& x^{2}-6 y=0\\
    \text{(k)}& 8 x^{2}+4 y=0\\
    \text{(l)}& 8 y^{2}+4 x=0
    \end{array}$

  2. Find the standard form of the equation of each parabola satisfying the given conditions.

    $\begin {array}{lll}
    \text{(a)}& \text{Focus}: \quad(7,0) ; & \text{Directrix}: \quad x=-7\\
    \text{(b)}& \text{Focus}: \quad(9,0) ; & \text{Directrix}: \quad x=-9\\
    \text{(c)}& \text{Focus}: \quad(-5,0) ; & \text{Directrix}: \quad x=5\\
    \text{(d)}& \text{Focus}: \quad(-10,0) ; & \text{Directrix}: \quad x=10\\
    \text{(e)}& \text{Focus}: \quad(0,15) ; & \text{Directrix}: \quad y=-15\\
    \text{(f)}& \text{Focus}: \quad(0,20) ; & \text{Directrix}: \quad y=-20\\
    \text{(g)}& \text{Focus}: \quad(0,-25) ; & \text{Directrix}: \quad y=25\\
    \text{(h)}& \text{Focus}: \quad(0,-15) ; & \text{Directrix}: \quad y=15\\
    \text{(i)}& \text{Vertex}:\quad(2,-3) ; & \text{Focus}: \quad(2,-5)\\
    \text{(j)}& \text{Vertex}:\quad(5,-2); & \text{Focus}: \quad(7,-2)\\
    \text{(k)}& \text{Focus}: \quad(3,2) ; & \text{Directrix}: \quad x=-1\\
    \text{(l)}& \text{Focus}: \quad(2,4) ; & \text{Directrix}: \quad x=-4\\
    \text{(m)}& \text{Focus}: \quad(-3,4) ; & \text{Directrix}: \quad y=2\\
    \text{(n)}& \text{Focus}: \quad(7,-1); & \text{Directrix}: \quad y=-9 \end{array}$

  3. Convert each equation to standard form and find the vertex, focus, and directrix of the parabola. Hence graph the parabola.

    $\begin{array}{ll} \text{(a)}& x^{2}-2 x-4 y+9=0\\ \text{(b)}& x^{2}+6 x+8 y+1=0\\ \text{(c)}& y^{2}-2 y+12 x-35=0\\ \text{(d)}& y^{2}-2 y-8 x+1=0\\ \text{(e)}& x^{2}+6 x-4 y+1=0\\ \text{(f)}& x^{2}+8 x-4 y+8=0 \end{array}$