Question

Find an equation of a parabola with a vertex at the origin and directrix X = 2.5

Answer 1

vertex=(0,0)
directrix is x=2.5
or x-2.5=0
so focus  S is (-2.5,0)
let P(x,y)be any point on the parabola.
Let M be the foot of perpendicular from P(x,y) on the directrix.
then SP=PM
$\sqrt{(x-(-2.5))^2+(y-0)^2} =| \frac{x-2.5}{( \sqrt{1})^2 } | squaring both sides$$(x+2.5)^2+y^2=(x-2.5)^2 , x^2+6.25+5x+y^2=x^2+6.25-5x$
$y^2=x^2+6.25-5x-x^2-6.25-5x=-10x so eq. is y^{2} =-10x$

Answer 2

To solve this problem, we just need to substitute the values into the equation and solve. The equation of the parabola is equal to y^2 = 10x

Equation of a Parabola

The standard equation of a parabola is given as

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

But in this question, the vertex is starting from the origin and the directrix is at 2.5

• vertex = (0,0)
• directrix x = 2.5
• h = 0
• k = 0
• a = -2.5

substituting the values into the equation;

$y^2 = 4*(-2.5)x\\ \\y^2 = 10x\\x = \frac{1}{10}y^2$

The equation of the parabola is equal to y^2 = 10x

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