Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, 9).

1 answer:

5 0

First we write standard form that includes both vertex and focus. The formula goes like this:

(x-h)^2 = 4*p*(y-k)

h and k are coordinates of vertex. since vertex is at beginning that means that h=k=0

p is focal lenght and in our case it is 9

expressing al of this we get:
x^2 = 4*9*y
x^2 = 36y
y = 1/36 * x^2

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Substitution: 45+15(6-1)

Simplify: 120

Step-by-step explanation:

Calculate within parenthases (6-1) =5 =45+15(5)

Multiply left to right 15(5)=75 =45+75

Add leftg to right 45+75=120

Answer = 120

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The length of a rectangle is 5 metres less than twice the breadth. If the perimeter is 50 meters,find the length and breadth

MAXImum [283]

As per the given question, it is stated that the length of a rectangle is 5 m less than twice the breadth.

Assumption : Let us assume the length as "l" and width as "b". So,

$\twoheadrightarrow \quad\sf{ Length =2(Width)-5}$

$\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}$

Also, we are given that the perimeter of the rectangle is 50 m. Basically, we need to apply here the formula of perimeter of rectangle which will act as a linear equation here.

$\\ \twoheadrightarrow \quad\sf{ Perimeter_{(Rectangle)} = 2(\ell +b) } \\$