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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{y^2} = 16x?

Explanation

Solution

Here in the given question, we are supposed to find the vertex, focus and directrix ofy2=16x{y^2} = 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{y^2} = 4ax, wherea0a \ne 0
The vertex of this parabola is at(0,0)\left( {0,0} \right)
The focus is at(a,0)\left( {a,0} \right)
The directrix is the line x+a=0x + a = 0

Complete step by step solution:
We are given the parabolic equation y2=16x{y^2} = 16x
The given parabola can be graphed as:

The given parabolic equation can very easily be written as y2=4(4)x{y^2} = 4\left( 4 \right)x
As we can clearly see that the given parabolic equation is in the form of y2=4ax{y^2} = 4axwhich is the general form of a parabola.
After comparing both the equations, we get a=4a = 4
Therefore, the vertex of given parabola
Vertex =(0,0)\left( {0,0} \right)
Similarly, the focus is
Focus =(a,0)\left( {a,0} \right)
Focus =(4,0)\left( {4,0} \right)
Similarly, the directrix for the parabola is
x+a=0 x+4=0 x=4  \Rightarrow x + a = 0 \\\ \Rightarrow x + 4 = 0 \\\ \Rightarrow x = - 4 \\\
Directrix =x=4x = - 4

Hence, the vertex is at(0,0)\left( {0,0} \right), the focus is at (4,0)\left( {4,0} \right) and the directrix is atx=4x = - 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.