Conic Section (Part 1) တွင် p>0 ဟုသတ်မှတ်၍ Parabola ကို x2=4py,x2=−4py,y2=4px နှင့် y2=−4px ဟူ ၍ ပုံသဏ္ဌာန် လေးမျိုးဖြင့် ဖော်ပြနိုင်ကြောင်း တင်ပြခဲ့ပြီး ဖြစ်သည်။ p ကို non-zero real number တစ်ခုအဖြစ် သတ်မှတ်ပြီး parabola ကို general form နှစ်ခုဖြင့်သာ မှတ်သားလေ့လာနိုင်ကြောင်း ယခု post တွင် ဆက်လက်တင်ပြပါမည်။
For y2=4px, p≠0,
When x=p,y2=4p2.
Therefore y=±2p.
Focus = (p,0)
The endpoints of latus rectum is (p,−2p) and (p,2p) .
And the length of latus rectum is |4p|.
For x2=4py, p≠0,
Focus = (0,p)
When y=p,x2=4p2.
Therefore x=±2p.
Hence the endpoints of latus rectum is (−2p,p) and (2p,p) .
And the length of latus rectum is |4p|.
Vertex က (0,0) ၌ ရှိသော Parabola ပုံစံလေးခု၏ အနှစ်ချုပ်ကို အောက်ပါဇယားဖြင့်မှတ်သားနိုင်ပါသည်။
Form (1)
- Equation : x2=4py
- Axis of Symmetry : y−axis
- Vertex at (0,0)
- Focus at (0,p)
- Directrix : y=−p
- If p>0, the parabola opens up
- If p<0, the parabola opens down
Form (2)
- Equation : y2=4px
- Axis of Symmetry : x−axis
- Vertex at (0,0)
- Focus at (p,0)
- Directrix : x=−p
- If p>0, the parabola opens to the right.
- If p<0, the parabola opens to the left.
Latus Rectum
The focal chord which is perpendicular to the axis is known as latus rectum of the conic. Axis of symmetry ပေါ် ထောင့်မတ်ကျသော Focal Chord ကို latus rectum ဟုခေါ်သည်။For y2=4px, p≠0,
When x=p,y2=4p2.
Therefore y=±2p.
Focus = (p,0)
The endpoints of latus rectum is (p,−2p) and (p,2p) .
And the length of latus rectum is |4p|.
For x2=4py, p≠0,
Focus = (0,p)
When y=p,x2=4p2.
Therefore x=±2p.
Hence the endpoints of latus rectum is (−2p,p) and (2p,p) .
And the length of latus rectum is |4p|.
Vertex က (0,0) ၌ ရှိသော Parabola ပုံစံလေးခု၏ အနှစ်ချုပ်ကို အောက်ပါဇယားဖြင့်မှတ်သားနိုင်ပါသည်။
Equation | Focus | Directrix | Axis of Symmetry | Endpoinds of Latus Rectum | Opens | |
---|---|---|---|---|---|---|
x2=4py | (0,p) | y=−p | y−axis | (±2p,p) | UP when p>0 | |
DOWN when p<0 | ||||||
y2=4px | (p,0) | x=−p | x−axis | (p,±2p) | RIGHT when p>0 | |
LEFT when p<0 |
Graphing Parabola
- to find the focus
- to find the equation of directrix
- to find the axis of symmetry
- to find the vertex
- to find the end points of latus rectum
- to draw the smooth curve passing through the vertex and the end points of latus rectum to form parabola
Example (1) Graph y2=24x. Solution Comparing with y2=4px, we get p=6 Axis of symmetry : x−axis The parabola opens to the right. Focus = (p,0)=(6,0) ∴ directrix : x=−6 Vertex = (0,0) The endpoints of latus rectum = (p,±2p)=(6,±12) Example (2) Graph x2=−6y. Solution Comparing with x2=4py, we get 4p=−6⇒p=−32 Axis of symmetry : y−axis The parabola opens down. Focus = (0,p)=(0,−32) ∴ directrix : y=32 Vertex = (0,0) The endpoints of latus rectum = (±2p, p)=(±3, −32) |
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