Question
Question: Find the vertex, focus and directrix of the parabolic equation \({y^2} = 16x\)?...
Find the vertex, focus and directrix of the parabolic equation y2=16x?
Solution
Here in the given question, we are supposed to find the vertex, focus and directrix ofy2=16x. In order to solve the given parabolic equation, we will compare it to the standard or general form of a parabolic equation. By comparing those we will get the value of vertex, focus and directrix of the parabola. Therefore, we will get the required result.
Formula used:
The general form of a parabolic equation is
y2=4ax, wherea=0
The vertex of this parabola is at(0,0)
The focus is at(a,0)
The directrix is the line x+a=0
Complete step by step solution:
We are given the parabolic equation y2=16x
The given parabola can be graphed as:
The given parabolic equation can very easily be written as y2=4(4)x
As we can clearly see that the given parabolic equation is in the form of y2=4axwhich is the general form of a parabola.
After comparing both the equations, we get a=4
Therefore, the vertex of given parabola
Vertex =(0,0)
Similarly, the focus is
Focus =(a,0)
Focus =(4,0)
Similarly, the directrix for the parabola is
⇒x+a=0 ⇒x+4=0 ⇒x=−4
Directrix =x=−4
Hence, the vertex is at(0,0), the focus is at (4,0) and the directrix is atx=−4.
Note: The given question was an easy one. Students should be aware of the parabolas and related concepts. The corresponding directrix is at infinity. A curve where any point is at an equal distance from a fixed point and a fixed straight line is known as a parabola. The formulas for the different parabolas are different. Students should not mix them up.