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Flashcards: Derive Parabola Equation: CCSS.Math.Content.HSG-GPE.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\textup{focus}= \left ( 1, 10\right )$ \(\displaystyle \textup{focus}= \left ( 1, 10\right )\)

$\textup{directrix } y = 7$ \(\displaystyle \textup{directrix } y = 7\)

$\left(x - 10\right)^{2} + \left(x - 1\right)^{2}$ \(\displaystyle \left(x - 10\right)^{2} + \left(x - 1\right)^{2}\)

$y = \frac{x^{2}}{12} - \frac{x}{6} + \frac{13}{3}$ \(\displaystyle y = \frac{x^{2}}{12} - \frac{x}{6} + \frac{13}{3}\)

$\left(x - 2\right)^{2}$ \(\displaystyle \left(x - 2\right)^{2}\)

$y = \frac{x^{2}}{24} - \frac{x}{12} + \frac{13}{6}$ \(\displaystyle y = \frac{x^{2}}{24} - \frac{x}{12} + \frac{13}{6}\)

$y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}$ \(\displaystyle y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}\)

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Flashcards: Derive Parabola Equation: CCSS.Math.Content.HSG-GPE.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\textup{focus}= \left ( 1, 10\right )$ \(\displaystyle \textup{focus}= \left ( 1, 10\right )\)

$\textup{directrix } y = 7$ \(\displaystyle \textup{directrix } y = 7\)

$\left(x - 10\right)^{2} + \left(x - 1\right)^{2}$ \(\displaystyle \left(x - 10\right)^{2} + \left(x - 1\right)^{2}\)

$y = \frac{x^{2}}{12} - \frac{x}{6} + \frac{13}{3}$ \(\displaystyle y = \frac{x^{2}}{12} - \frac{x}{6} + \frac{13}{3}\)

$\left(x - 2\right)^{2}$ \(\displaystyle \left(x - 2\right)^{2}\)

$y = \frac{x^{2}}{24} - \frac{x}{12} + \frac{13}{6}$ \(\displaystyle y = \frac{x^{2}}{24} - \frac{x}{12} + \frac{13}{6}\)

$y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}$ \(\displaystyle y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}\)

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Flashcards: Derive Parabola Equation: CCSS.Math.Content.HSG-GPE.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\textup{focus}= \left ( 1, 10\right )$ \(\displaystyle \textup{focus}= \left ( 1, 10\right )\)

$\textup{directrix } y = 7$ \(\displaystyle \textup{directrix } y = 7\)

Your answer:

$\left(x - 10\right)^{2} + \left(x - 1\right)^{2}$ \(\displaystyle \left(x - 10\right)^{2} + \left(x - 1\right)^{2}\)

Your answer:

$y = \frac{x^{2}}{12} - \frac{x}{6} + \frac{13}{3}$ \(\displaystyle y = \frac{x^{2}}{12} - \frac{x}{6} + \frac{13}{3}\)

Your answer:

$\left(x - 2\right)^{2}$ \(\displaystyle \left(x - 2\right)^{2}\)

Your answer:

$y = \frac{x^{2}}{24} - \frac{x}{12} + \frac{13}{6}$ \(\displaystyle y = \frac{x^{2}}{24} - \frac{x}{12} + \frac{13}{6}\)

Correct answer:

$y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}$ \(\displaystyle y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}\)

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$ \(\displaystyle \left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}\)

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$ \(\displaystyle \left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}\)

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$ \(\displaystyle y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}\)

Now we can substitute 1 for a 10 for b and 7 for y

$y_{0}^{2} - 14 y_{0} + 49 = x_{0}^{2} - 2 x_{0} + y_{0}^{2} - 20 y_{0} + 101$ \(\displaystyle y_{0}^{2} - 14 y_{0} + 49 = x_{0}^{2} - 2 x_{0} + y_{0}^{2} - 20 y_{0} + 101\)

Now we can simplify, and solve for $y_{0}$ \(\displaystyle y_{0}\)

$y_{0} = \frac{x_{0}^{2}}{6} - \frac{x_{0}}{3} + \frac{26}{3}$ \(\displaystyle y_{0} = \frac{x_{0}^{2}}{6} - \frac{x_{0}}{3} + \frac{26}{3}\)

So our answer is then

$y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}$ \(\displaystyle y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}\)

Flashcards: Derive Parabola Equation: CCSS.Math.Content.HSG-GPE.A.2

Find the parabolic equation, where the focus and the directrix are as follows.

$\\\text{focus}= \left ( 1, 10\right ) \\\text{directrix } y = 7$ \(\displaystyle \\\text{focus}= \left ( 1, 10\right ) \\\text{directrix } y = 7\)

$y = \frac{x^{2}}{12} - \frac{x}{6} + \frac{13}{3}$ \(\displaystyle y = \frac{x^{2}}{12} - \frac{x}{6} + \frac{13}{3}\)

$\left(x - 10\right)^{2} + \left(x - 1\right)^{2}$ \(\displaystyle \left(x - 10\right)^{2} + \left(x - 1\right)^{2}\)

$y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}$ \(\displaystyle y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}\)

$y = \frac{x^{2}}{24} - \frac{x}{12} + \frac{13}{6}$ \(\displaystyle y = \frac{x^{2}}{24} - \frac{x}{12} + \frac{13}{6}\)

$\left(x - 2\right)^{2}$ \(\displaystyle \left(x - 2\right)^{2}\)

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Flashcards: Derive Parabola Equation: CCSS.Math.Content.HSG-GPE.A.2

Find the parabolic equation, where the focus and the directrix are as follows.

$\\\text{focus}= \left ( 1, 10\right ) \\\text{directrix } y = 7$ \(\displaystyle \\\text{focus}= \left ( 1, 10\right ) \\\text{directrix } y = 7\)

Your answer:

$y = \frac{x^{2}}{12} - \frac{x}{6} + \frac{13}{3}$ \(\displaystyle y = \frac{x^{2}}{12} - \frac{x}{6} + \frac{13}{3}\)

Your answer:

$\left(x - 10\right)^{2} + \left(x - 1\right)^{2}$ \(\displaystyle \left(x - 10\right)^{2} + \left(x - 1\right)^{2}\)

Your answer:

$y = \frac{x^{2}}{24} - \frac{x}{12} + \frac{13}{6}$ \(\displaystyle y = \frac{x^{2}}{24} - \frac{x}{12} + \frac{13}{6}\)

Your answer:

$\left(x - 2\right)^{2}$ \(\displaystyle \left(x - 2\right)^{2}\)

Correct answer:

$y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}$ \(\displaystyle y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}\)

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$ \(\displaystyle \left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}\)

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$ \(\displaystyle \left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}\)

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$ \(\displaystyle y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}\)

Now we can substitute 1 for a 10 for b and 7 for y

$y_{0}^{2} - 14 y_{0} + 49 = x_{0}^{2} - 2 x_{0} + y_{0}^{2} - 20 y_{0} + 101$ \(\displaystyle y_{0}^{2} - 14 y_{0} + 49 = x_{0}^{2} - 2 x_{0} + y_{0}^{2} - 20 y_{0} + 101\)

Now we can simplify, and solve for $y_{0}$ \(\displaystyle y_{0}\)

$y_{0} = \frac{x_{0}^{2}}{6} - \frac{x_{0}}{3} + \frac{26}{3}$ \(\displaystyle y_{0} = \frac{x_{0}^{2}}{6} - \frac{x_{0}}{3} + \frac{26}{3}\)

So our answer is then

$y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}$ \(\displaystyle y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}\)

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