Introduction to Plug in and Solve Parabolas:
A curve which is formed by the intersection of a right circular cone and half of the circle is called as parabola. Directrix of a parabola is a set of all points located at same distance from a fixed line. The fixed point is called as focus and not on the directrix. The midpoint between focus of a parabola and vertex of a parabola is called as vertex. A line passing through vertex and focus of a parabola is called as axis of symmetry. Finding the above four criteria by solving parabola equation. Let us see about plug in and solve parabolas in this article.
Worked Examples to Plug in and Solve Parabolas
The general form the parabolic curve is` y = ax^(2) + bx + c` or `y^(2) = 4ax` . Substitute the above formula to find the vertices, latus rectum, focus and axis of symmetry.
Example 1 for Plug in and Solve Parabolas – Vertex:
Find the vertex of a parabola equation `y = x^ (2) + 4x + 3` .
Solution:
Given parabola equation is `y = x^ (2) + 4x + 3` .
To find the vertices of a given parabola, we have to plug `y = 0` in the above equation, we get,
` 0 = x^ (2) + 4x + 3`
Now we have to factor the above equation, we get,
So `x^ (2) + x + 3x + 3 = 0`
`x(x + 1) + 3 (x + 1) = 0`
`(x + 1) (x + 3) = 0`
From this `x + 1 = 0` and `x + 3 = 0`
Then `x = - 1` and `x = - 3`
So, the vertices of given parabola equation is` (-1, 0)` and `(-3, 0)` .
Example 2 for Plug in and Solve Parabolas – Focus:
What is the focus of the following parabola equation `y^ (2) = 8x` ?
Solution:
Given parabola equation is `y^ (2) = 8x` is of the form y`^ (2) = 4ax`
We know that the formula for focus, `p = 1 / (4a)`
Now compare the given equation y2 = 8x with the general equation `y^ (2) = 4ax` . So, that `4a = 8`
From this,` p = 1 / (4a) = 1 / 8`
So, the focus of a parabola equation is `(0, 1/8)` .
Other Example Problems to Plug in and Solve Parabolas
Example 3 for Plug in and Solve Parabolas – Axis of Symmetry:
What is the axis of symmetry for parabola equation `y = 5x^ (2) + 15x + 12` `?`
Solution:
Given parabola equation is `y = 5x^ (2) + 15x + 12`
From the above equation, plug `a = 5` and `b = 15` in the axis formula.
So the axis of the symmetry of the given parabola is `-b/ (2a) = - 15/ (2 xx 5) = -15/10 = - 3/2`
Therefore, the axis of symmetry for a given parabolic curve equation is` -3/2` .
Example 4 for Plug in and Solve Parabolas – Latus Rectum:
Find the latus rectum of the given parabola equation `y^ (2) = 12x` .
Solution:
The given parabola equation is `y^ (2) = 12x`
To find the latus rectum, we have to find the value of` p` .
The parabola equation is of the form `y^ (2) = 4ax`
Here `4a = 12`
So, `p = 1/ (4a) = 1/12`
The formula for latus rectum is `4p` .
Plug the value for `p =1/12` in the latus rectum formula.
From this, the latus rectum of the parabola is `= 4p = 4 (1/12) = 4/12 = 1/3`
Therefore, the latus rectum for the parabola equation is `1/3` .
A curve which is formed by the intersection of a right circular cone and half of the circle is called as parabola. Directrix of a parabola is a set of all points located at same distance from a fixed line. The fixed point is called as focus and not on the directrix. The midpoint between focus of a parabola and vertex of a parabola is called as vertex. A line passing through vertex and focus of a parabola is called as axis of symmetry. Finding the above four criteria by solving parabola equation. Let us see about plug in and solve parabolas in this article.
Worked Examples to Plug in and Solve Parabolas
The general form the parabolic curve is` y = ax^(2) + bx + c` or `y^(2) = 4ax` . Substitute the above formula to find the vertices, latus rectum, focus and axis of symmetry.
Example 1 for Plug in and Solve Parabolas – Vertex:
Find the vertex of a parabola equation `y = x^ (2) + 4x + 3` .
Solution:
Given parabola equation is `y = x^ (2) + 4x + 3` .
To find the vertices of a given parabola, we have to plug `y = 0` in the above equation, we get,
` 0 = x^ (2) + 4x + 3`
Now we have to factor the above equation, we get,
So `x^ (2) + x + 3x + 3 = 0`
`x(x + 1) + 3 (x + 1) = 0`
`(x + 1) (x + 3) = 0`
From this `x + 1 = 0` and `x + 3 = 0`
Then `x = - 1` and `x = - 3`
So, the vertices of given parabola equation is` (-1, 0)` and `(-3, 0)` .
Example 2 for Plug in and Solve Parabolas – Focus:
What is the focus of the following parabola equation `y^ (2) = 8x` ?
Solution:
Given parabola equation is `y^ (2) = 8x` is of the form y`^ (2) = 4ax`
We know that the formula for focus, `p = 1 / (4a)`
Now compare the given equation y2 = 8x with the general equation `y^ (2) = 4ax` . So, that `4a = 8`
From this,` p = 1 / (4a) = 1 / 8`
So, the focus of a parabola equation is `(0, 1/8)` .
Other Example Problems to Plug in and Solve Parabolas
Example 3 for Plug in and Solve Parabolas – Axis of Symmetry:
What is the axis of symmetry for parabola equation `y = 5x^ (2) + 15x + 12` `?`
Solution:
Given parabola equation is `y = 5x^ (2) + 15x + 12`
From the above equation, plug `a = 5` and `b = 15` in the axis formula.
So the axis of the symmetry of the given parabola is `-b/ (2a) = - 15/ (2 xx 5) = -15/10 = - 3/2`
Therefore, the axis of symmetry for a given parabolic curve equation is` -3/2` .
Example 4 for Plug in and Solve Parabolas – Latus Rectum:
Find the latus rectum of the given parabola equation `y^ (2) = 12x` .
Solution:
The given parabola equation is `y^ (2) = 12x`
To find the latus rectum, we have to find the value of` p` .
The parabola equation is of the form `y^ (2) = 4ax`
Here `4a = 12`
So, `p = 1/ (4a) = 1/12`
The formula for latus rectum is `4p` .
Plug the value for `p =1/12` in the latus rectum formula.
From this, the latus rectum of the parabola is `= 4p = 4 (1/12) = 4/12 = 1/3`
Therefore, the latus rectum for the parabola equation is `1/3` .
