Math / Geometry

QuestionIf the coefficient $a$ in the general quadratic equation is equal to $\frac{1}{48}$ , what is the focal length for the parabola, $c^{\prime}$ , equal to? 12 1 $\frac{1}{48}$ 48 $\frac{1}{12}$ 4 $\frac{1}{24}$

Studdy Solution

STEP 1

1. The quadratic equation is in the form $y = ax^2$ .
2. The parabola is oriented such that its vertex is at the origin.
3. The focal length $c^{\prime}$ is the distance from the vertex to the focus of the parabola.

STEP 2

1. Recall the relationship between the coefficient $a$ and the focal length $c^{\prime}$ for a parabola.
2. Substitute the given value of $a$ .
3. Solve for the focal length $c^{\prime}$ .

STEP 3

Recall the relationship between the coefficient $a$ and the focal length $c^{\prime}$ for a parabola:
For a parabola $y = ax^2$ , the focal length $c^{\prime}$ is given by:
$c^{\prime} = \frac{1}{4a}$

STEP 4

Substitute the given value of $a = \frac{1}{48}$ into the formula:
$c^{\prime} = \frac{1}{4 \times \frac{1}{48}}$

STEP 5

Solve for the focal length $c^{\prime}$ :
$c^{\prime} = \frac{1}{\frac{4}{48}}$ $c^{\prime} = \frac{48}{4}$ $c^{\prime} = 12$
The focal length $c^{\prime}$ is:
$\boxed{12}$

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QuestionIf the coefficient $a$ in the general quadratic equation is equal to $\frac{1}{48}$ , what is the focal length for the parabola, $c^{\prime}$ , equal to? 12 1 $\frac{1}{48}$ 48 $\frac{1}{12}$ 4 $\frac{1}{24}$

Studdy Solution

STEP 1

1. The quadratic equation is in the form $y = ax^2$ .
2. The parabola is oriented such that its vertex is at the origin.
3. The focal length $c^{\prime}$ is the distance from the vertex to the focus of the parabola.

STEP 2

1. Recall the relationship between the coefficient $a$ and the focal length $c^{\prime}$ for a parabola.
2. Substitute the given value of $a$ .
3. Solve for the focal length $c^{\prime}$ .

STEP 3

Recall the relationship between the coefficient $a$ and the focal length $c^{\prime}$ for a parabola:
For a parabola $y = ax^2$ , the focal length $c^{\prime}$ is given by:
$c^{\prime} = \frac{1}{4a}$

STEP 4

Substitute the given value of $a = \frac{1}{48}$ into the formula:
$c^{\prime} = \frac{1}{4 \times \frac{1}{48}}$

STEP 5

Solve for the focal length $c^{\prime}$ :
$c^{\prime} = \frac{1}{\frac{4}{48}}$ $c^{\prime} = \frac{48}{4}$ $c^{\prime} = 12$
The focal length $c^{\prime}$ is:
$\boxed{12}$

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QuestionIf the coefficient aaa in the general quadratic equation is equal to 148\frac{1}{48}481​, what is the focal length for the parabola, c′c^{\prime}c′, equal to? 12 1 148\frac{1}{48}481​ 48 112\frac{1}{12}121​ 4 124\frac{1}{24}241​

Studdy Solution

STEP 1

1. The quadratic equation is in the form y=ax2 y = ax^2 y=ax2.2. The parabola is oriented such that its vertex is at the origin.3. The focal length c′ c^{\prime} c′ is the distance from the vertex to the focus of the parabola.

STEP 2

1. Recall the relationship between the coefficient a a a and the focal length c′ c^{\prime} c′ for a parabola.2. Substitute the given value of a a a.3. Solve for the focal length c′ c^{\prime} c′.

STEP 3

Recall the relationship between the coefficient a a a and the focal length c′ c^{\prime} c′ for a parabola:For a parabola y=ax2 y = ax^2 y=ax2, the focal length c′ c^{\prime} c′ is given by:c′=14a c^{\prime} = \frac{1}{4a} c′=4a1​

STEP 4

Substitute the given value of a=148 a = \frac{1}{48} a=481​ into the formula:c′=14×148 c^{\prime} = \frac{1}{4 \times \frac{1}{48}} c′=4×481​1​

STEP 5

Solve for the focal length c′ c^{\prime} c′:c′=1448 c^{\prime} = \frac{1}{\frac{4}{48}} c′=484​1​ c′=484 c^{\prime} = \frac{48}{4} c′=448​ c′=12 c^{\prime} = 12 c′=12 The focal length c′ c^{\prime} c′ is:12 \boxed{12} 12​

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QuestionIf the coefficient $a$ in the general quadratic equation is equal to $\frac{1}{48}$ , what is the focal length for the parabola, $c^{\prime}$ , equal to? 12 1 $\frac{1}{48}$ 48 $\frac{1}{12}$ 4 $\frac{1}{24}$

1. The quadratic equation is in the form $y = ax^2$ .
2. The parabola is oriented such that its vertex is at the origin.
3. The focal length $c^{\prime}$ is the distance from the vertex to the focus of the parabola.

1. Recall the relationship between the coefficient $a$ and the focal length $c^{\prime}$ for a parabola.
2. Substitute the given value of $a$ .
3. Solve for the focal length $c^{\prime}$ .

Recall the relationship between the coefficient $a$ and the focal length $c^{\prime}$ for a parabola:
For a parabola $y = ax^2$ , the focal length $c^{\prime}$ is given by:
$c^{\prime} = \frac{1}{4a}$

Substitute the given value of $a = \frac{1}{48}$ into the formula:
$c^{\prime} = \frac{1}{4 \times \frac{1}{48}}$

Solve for the focal length $c^{\prime}$ :
$c^{\prime} = \frac{1}{\frac{4}{48}}$ $c^{\prime} = \frac{48}{4}$ $c^{\prime} = 12$
The focal length $c^{\prime}$ is:
$\boxed{12}$