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Common Core: High School - Geometry : Derive Parabola Equation: CCSS.Math.Content.HSG-GPE.A.2

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Example Questions

Example Question #11 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

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

$\\\text{focus}= \left ( 2, 5\right ) \\\text{directrix } y = -6$

Possible Answers:

$y = \frac{x^{2}}{44} - \frac{x}{11} - \frac{7}{44}$

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

$\left(x - 5\right)^{2} + \left(x - 2\right)^{2}$

$y = \frac{x^{2}}{22} - \frac{2 x}{11} - \frac{7}{22}$

$y = \frac{x^{2}}{88} - \frac{x}{22} - \frac{7}{88}$

Correct answer:

$y = \frac{x^{2}}{22} - \frac{2 x}{11} - \frac{7}{22}$

Explanation:

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

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

Let's square each side

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

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute 2 for a 5 for b and -6 for y

$y_{0}^{2} + 12 y_{0} + 36 = x_{0}^{2} - 4 x_{0} + y_{0}^{2} - 10 y_{0} + 29$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{22} - \frac{2 x_{0}}{11} - \frac{7}{22}$

So our answer is then

$y = \frac{x^{2}}{22} - \frac{2 x}{11} - \frac{7}{22}$

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Example Question #12 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

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

$\\\text{focus}= \left ( -8, 9\right ) \\\text{directrix } y = 12$

Possible Answers:

$y = - \frac{x^{2}}{24} - \frac{2 x}{3} - \frac{1}{24}$

$y = - \frac{x^{2}}{12} - \frac{4 x}{3} - \frac{1}{12}$

$\left(x - 9\right)^{2} + \left(x + 8\right)^{2}$

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

$y = - \frac{x^{2}}{6} - \frac{8 x}{3} - \frac{1}{6}$

Correct answer:

$y = - \frac{x^{2}}{6} - \frac{8 x}{3} - \frac{1}{6}$

Explanation:

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

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

Let's square each side

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

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute -8 for a 9 for b and 12 for y

$y_{0}^{2} - 24 y_{0} + 144 = x_{0}^{2} + 16 x_{0} + y_{0}^{2} - 18 y_{0} + 145$

Now we can simplify, and solve for $y_{0}$

$y_{0} = - \frac{x_{0}^{2}}{6} - \frac{8 x_{0}}{3} - \frac{1}{6}$

So our answer is then

$y = - \frac{x^{2}}{6} - \frac{8 x}{3} - \frac{1}{6}$

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Example Question #11 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

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

$\\\text{focus}= \left ( -6, 1\right ) \\\text{directrix } y = -5$

Possible Answers:

$y = \frac{x^{2}}{12} + x + 1$

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

$y = \frac{x^{2}}{48} + \frac{x}{4} + \frac{1}{4}$

$y = \frac{x^{2}}{24} + \frac{x}{2} + \frac{1}{2}$

$\left(x - 1\right)^{2} + \left(x + 6\right)^{2}$

Correct answer:

$y = \frac{x^{2}}{12} + x + 1$

Explanation:

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

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

Let's square each side

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

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute -6 for a 1 for b and -5 for y

$y_{0}^{2} + 10 y_{0} + 25 = x_{0}^{2} + 12 x_{0} + y_{0}^{2} - 2 y_{0} + 37$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{12} + x_{0} + 1$

So our answer is then

$y = \frac{x^{2}}{12} + x + 1$

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Example Question #14 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

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

$\\\text{focus}= \left ( -6, 6\right ) \\\text{directrix } y = -6$

Possible Answers:

$y = \frac{x^{2}}{96} + \frac{x}{8} + \frac{3}{8}$

$y = \frac{x^{2}}{24} + \frac{x}{2} + \frac{3}{2}$

$y = \frac{x^{2}}{48} + \frac{x}{4} + \frac{3}{4}$

$\left(x - 6\right)^{2} + \left(x + 6\right)^{2}$

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

Correct answer:

$y = \frac{x^{2}}{24} + \frac{x}{2} + \frac{3}{2}$

Explanation:

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

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

Let's square each side

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

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute -6 for a 6 for b and -6 for y

$y_{0}^{2} + 12 y_{0} + 36 = x_{0}^{2} + 12 x_{0} + y_{0}^{2} - 12 y_{0} + 72$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{24} + \frac{x_{0}}{2} + \frac{3}{2}$

So our answer is then

$y = \frac{x^{2}}{24} + \frac{x}{2} + \frac{3}{2}$

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Common Core: High School - Geometry : Derive Parabola Equation: CCSS.Math.Content.HSG-GPE.A.2

Study concepts, example questions & explanations for Common Core: High School - Geometry

All Common Core: High School - Geometry Resources

Example Questions

Example Question #11 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #12 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #11 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #14 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

All Common Core: High School - Geometry Resources

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