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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 #1 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

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

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

$\textup{directrix } y = 7$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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 1 for a 10 for b and 7 for y

$y_{0}^{2} - 14 y_{0} + 49 = x_{0}^{2} - 2 x_{0} + y_{0}^{2} - 20 y_{0} + 101$

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

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

So our answer is then

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

Report an Error

Example Question #12 : Expressing Geometric Properties With Equations

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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 1 for a 10 for b and 7 for y

$y_{0}^{2} - 14 y_{0} + 49 = x_{0}^{2} - 2 x_{0} + y_{0}^{2} - 20 y_{0} + 101$

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

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

So our answer is then

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

Report an Error

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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 1 for a 10 for b and 7 for y

$y_{0}^{2} - 14 y_{0} + 49 = x_{0}^{2} - 2 x_{0} + y_{0}^{2} - 20 y_{0} + 101$

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

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

So our answer is then

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

Report an Error

Example Question #3 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

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

$\\\text{focus}= \left ( -10, 4\right ) \\ \text{directrix } y = -11$

Possible Answers:

$y = \frac{x^{2}}{60} + \frac{x}{3} - \frac{1}{12}$

$\left(x - 4\right)^{2} + \left(x + 10\right)^{2}$

$y = \frac{x^{2}}{120} + \frac{x}{6} - \frac{1}{24}$

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

$y = \frac{x^{2}}{30} + \frac{2 x}{3} - \frac{1}{6}$

Correct answer:

$y = \frac{x^{2}}{30} + \frac{2 x}{3} - \frac{1}{6}$

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 -10 for a 4 for b and -11 for y

$y_{0}^{2} + 22 y_{0} + 121 = x_{0}^{2} + 20 x_{0} + y_{0}^{2} - 8 y_{0} + 116$

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

$y_{0} = \frac{x_{0}^{2}}{30} + \frac{2 x_{0}}{3} - \frac{1}{6}$

So our answer is then

$y = \frac{x^{2}}{30} + \frac{2 x}{3} - \frac{1}{6}$

Report an Error

Example Question #1 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

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

$\\\text{focus}= \left ( 6, -9\right ) \\ \text{directrix } y = -5$

Possible Answers:

$y = - \frac{x^{2}}{8} + \frac{3 x}{2} - \frac{23}{2}$

$\left(x - 6\right)^{2} + \left(x + 9\right)^{2}$

$y = - \frac{x^{2}}{16} + \frac{3 x}{4} - \frac{23}{4}$

$y = - \frac{x^{2}}{32} + \frac{3 x}{8} - \frac{23}{8}$

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

Correct answer:

$y = - \frac{x^{2}}{8} + \frac{3 x}{2} - \frac{23}{2}$

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 -9 for b and -5 for y

$y_{0}^{2} + 10 y_{0} + 25 = x_{0}^{2} - 12 x_{0} + y_{0}^{2} + 18 y_{0} + 117$

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

$y_{0} = - \frac{x_{0}^{2}}{8} + \frac{3 x_{0}}{2} - \frac{23}{2}$

So our answer is then

$y = - \frac{x^{2}}{8} + \frac{3 x}{2} - \frac{23}{2}$

Report an Error

Example Question #6 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

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

$\\\text{focus}= \left ( 1, -6\right ) \\\text{directrix} y = -19$

Possible Answers:

$y = \frac{x^{2}}{52} - \frac{x}{26} - \frac{81}{13}$

$y = \frac{x^{2}}{104} - \frac{x}{52} - \frac{81}{26}$

$y = \frac{x^{2}}{26} - \frac{x}{13} - \frac{162}{13}$

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

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

Correct answer:

$y = \frac{x^{2}}{26} - \frac{x}{13} - \frac{162}{13}$

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 1 for a -6 for b and -19 for y

$y_{0}^{2} + 38 y_{0} + 361 = x_{0}^{2} - 2 x_{0} + y_{0}^{2} + 12 y_{0} + 37$

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

$y_{0} = \frac{x_{0}^{2}}{26} - \frac{x_{0}}{13} - \frac{162}{13}$

So our answer is then

$y = \frac{x^{2}}{26} - \frac{x}{13} - \frac{162}{13}$

Report an Error

Example Question #13 : Expressing Geometric Properties With Equations

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

$\\\text{focus}= \left ( -10, 6\right ) \\ \text{directrix } y = 15$

Possible Answers:

$y = - \frac{x^{2}}{72} - \frac{5 x}{18} + \frac{89}{72}$

$\left(x - 6\right)^{2} + \left(x + 10\right)^{2}$

$y = - \frac{x^{2}}{18} - \frac{10 x}{9} + \frac{89}{18}$

$y = - \frac{x^{2}}{36} - \frac{5 x}{9} + \frac{89}{36}$

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

Correct answer:

$y = - \frac{x^{2}}{18} - \frac{10 x}{9} + \frac{89}{18}$

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 -10 for a 6 for b and 15 for y

$y_{0}^{2} - 30 y_{0} + 225 = x_{0}^{2} + 20 x_{0} + y_{0}^{2} - 12 y_{0} + 136$

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

$y_{0} = - \frac{x_{0}^{2}}{18} - \frac{10 x_{0}}{9} + \frac{89}{18}$

So our answer is then

$y = - \frac{x^{2}}{18} - \frac{10 x}{9} + \frac{89}{18}$

Report an Error

Example Question #14 : Expressing Geometric Properties With Equations

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

$\\\text{focus}= \left ( 7, 5\right ) \\\text{directrix } y = -4$

Possible Answers:

$y = \frac{x^{2}}{72} - \frac{7 x}{36} + \frac{29}{36}$

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

$y = \frac{x^{2}}{36} - \frac{7 x}{18} + \frac{29}{18}$

$\left(x - 7\right)^{2} + \left(x - 5\right)^{2}$

$y = \frac{x^{2}}{18} - \frac{7 x}{9} + \frac{29}{9}$

Correct answer:

$y = \frac{x^{2}}{18} - \frac{7 x}{9} + \frac{29}{9}$

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 7 for a 5 for b and -4 for y

$y_{0}^{2} + 8 y_{0} + 16 = x_{0}^{2} - 14 x_{0} + y_{0}^{2} - 10 y_{0} + 74$

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

$y_{0} = \frac{x_{0}^{2}}{18} - \frac{7 x_{0}}{9} + \frac{29}{9}$

So our answer is then

$y = \frac{x^{2}}{18} - \frac{7 x}{9} + \frac{29}{9}$

Report an Error

Example Question #21 : Expressing Geometric Properties With Equations

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

$\\\text{focus}= \left ( 6, 8\right ) \\ \text{directrix } y = 10$

Possible Answers:

$\left(x - 8\right)^{2} + \left(x - 6\right)^{2}$

$y = \frac{x}{8} \left(- x + 12\right)$

$y = \frac{x}{16} \left(- x + 12\right)$

$y = \frac{x}{4} \left(- x + 12\right)$

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

Correct answer:

$y = \frac{x}{4} \left(- x + 12\right)$

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 8 for b and 10 for y

$y_{0}^{2} - 20 y_{0} + 100 = x_{0}^{2} - 12 x_{0} + y_{0}^{2} - 16 y_{0} + 100$

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

$y_{0} = \frac{x_{0}}{4} \left(- x_{0} + 12\right)$

So our answer is then

$y = \frac{x}{4} \left(- x + 12\right)$

Report an Error

Example Question #9 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

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

$\\\text{focus}= \left ( -10, -3\right ) \\\text{directrix } y = -4$

Possible Answers:

$y = \frac{x^{2}}{8} + \frac{5 x}{2} + \frac{93}{8}$

$y = \frac{x^{2}}{2} + 10 x + \frac{93}{2}$

$\left(x + 3\right)^{2} + \left(x + 10\right)^{2}$

$y = \frac{x^{2}}{4} + 5 x + \frac{93}{4}$

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

Correct answer:

$y = \frac{x^{2}}{2} + 10 x + \frac{93}{2}$

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 -10 for a -3 for b and -4 for y

$y_{0}^{2} + 8 y_{0} + 16 = x_{0}^{2} + 20 x_{0} + y_{0}^{2} + 6 y_{0} + 109$

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

$y_{0} = \frac{x_{0}^{2}}{2} + 10 x_{0} + \frac{93}{2}$

So our answer is then

$y = \frac{x^{2}}{2} + 10 x + \frac{93}{2}$

Report an Error

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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 #1 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #12 : Expressing Geometric Properties With Equations

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

Example Question #3 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #1 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #6 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #13 : Expressing Geometric Properties With Equations

Example Question #14 : Expressing Geometric Properties With Equations

Example Question #21 : Expressing Geometric Properties With Equations

Example Question #9 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

All Common Core: High School - Geometry Resources

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