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Find the standard form of the equation of the parabola with the given characteristic(s) and vertex at the origin

Sagot :

The standard form of the equation of the parabola with the given characteristic(s) and vertex at the origin and directrix is y² = 16x.

What is the parabola?

It is the locus of a point that moves so that it is always the same distance from a non-movable point and a given line. The non-movable point is called focus and the non-movable line is called the directrix.

Given that the parabola's vertex is located at the origin and that the directrix is at x = -4, the focus is at (4,0).

Then the equation can be written as

[tex]\rm \sqrt{(x-4)^2 +(y-0)^2} = \sqrt{(x+4)^2 + (y-y)^2}\\\\\sqrt{(x-4)^2 +(y-0)^2} = \sqrt{(x+4)^2}[/tex]

On squaring both sides, we have

[tex]\begin{aligned} (x-4)^2 +(y-0)^2 &= (x+4)^2 \\\\x^2 + 16 - 8x +y^2 &= x^2 + 16 + 8x\\\\y^2 &= 16x \end{aligned}[/tex]

More about the parabola link is given below.

https://brainly.com/question/8495504

#SPJ4

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