Question

Help, please!!

A parabola passes through the point (5,6) and has a vertex at (3,4). What's its equation??

Answer 1

To find the equation of the parabola, we need to use the standard form of the equation for a parabola, which is y = a(x-h)^2 + k, where (h,k) is the vertex.

Given that the vertex is (3,4), we can substitute these values into the equation to get y = a(x-3)^2 + 4.

Now, let's use the fact that the parabola passes through the point (5,6). We can substitute these values into the equation to get 6 = a(5-3)^2 + 4.

Simplifying the equation, we have 6 = 4a + 4.

Subtracting 4 from both sides, we get 2 = 4a.

Dividing both sides by 4, we find that a = 1/2.

Now we can substitute the value of a back into the equation to find the final equation of the parabola. Plugging in a = 1/2, we have y = (1/2)(x-3)^2 + 4.

So, the equation of the parabola is y = (1/2)(x-3)^2 + 4.

Explanation:

<3

Answer 2

`\textsf{Vertical axis:} \quad y = (1)/(2)(x - 3)^2 + 4`

`\textsf{Horizontal axis:}\quad (y - 4)^2 = 2(x - 3)`

Explanation:

The standard equation for a parabola with a vertical axis is:

`\large\boxed{y = a(x-h)^2 + k}`

where:

• (h, k) is the vertex.
• "a" is a constant that determines the direction and steepness of the parabola.

Given the vertex is (3, 4), then:

• h = 3
• k = 4

Substitute the vertex, along with the given point (5, 6) through which the parabola passes, into the general equation, and solve for "a".

`\begin{aligned}y &= a(x-h)^2 + k\\\\\implies 6&=a(5-3)^2+4\\\\6&=a(2)^2+4\\\\6&=4a+4\\\\6-4&=4a+4-4\\\\2&=4a\\\\(2)/(4)&=(4a)/(4)\\\\(1)/(2)&=a\end{aligned}`

Substitute the found value of a and the vertex into the general equation to write the equation of the parabola (with a vertical axis):

`\large\boxed{\boxed{y = (1)/(2)(x - 3)^2 + 4}}`

`\hrulefill`

We have assumed that the parabola has a vertical axis (i.e. it opens up or down). However, if the parabola has a horizontal axis (opens left or right), the standard form of the equation for this type of parabola is:

`\large\boxed{(y - k)^2 = 4p(x - h)}`

where:

• (h, k) is the vertex.
• "p" is the distance between the vertex and the focus (or between the vertex and the directrix).

Substitute the vertex (3, 4) and the point (5, 6) into the equation and solve for p:

`\begin{aligned}(y-k)^2&=4p(x-h)\\\\\implies (6-4)^2&=4p(5-3)\\\\2^2&=4p(2)\\\\4&=8p\\\\(4)/(8)&=(8p)/(8)\\\\(1)/(2)&=p\end{aligned}`

Substitute the found value of p and the vertex into the general equation to write the equation of the parabola (with a horizontal axis):

`\large\boxed{\boxed{(y - 4)^2 = 2(x - 3)}}`