Common Core Standard HSG-GPE.A.2 Questions

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Grade 11 Quadratic Equations and Expressions CCSS: HSG-GPE.A.2

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Which of the following equations is correct, and why?

$d_1^2 + d_2^2 = x^2$ , by the Pythagorean Theorem.
$d_1+d_2=0$ , because the distances will always be opposite.
$d_1 = d_2$ , by definition of a parabola.
$k*d_1 = h*d_2$ , because the ratio $k/h$ is always constant.

Grade 11 Quadratic Equations and Expressions CCSS: HSG-GPE.A.2

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Is the answer from the previous question the correct equation for the parabola with focus at

$(6,2)$ and directrix

$x=4 ?$ Why or why not?

No, the vertex has not yet been found or given, and without this information one cannot find the equation.
No, since there is a $y^2$ term, but no $x^2$ term.
Yes, but the translation from the origin still needs to be accounted for.
Yes, it's just in a different form (and can be put in the more common form by completing the square).

Grade 11 Quadratic Equations and Expressions CCSS: HSG-GPE.A.2

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After substituting in the equations for

$d_1, d_2$ found in the second and third questions into the equation in the previous question, which of the following is the resulting equation? (Fully expand and put all terms onto one side.)

$2x^2 + y^2 - 20x - 4y + 56 = 0$
$x^2 + y^2 - 20x - 4y + 56 = 0$
$y - x^2 + 6x + 2 = 0$
$y^2 - 4x - 4y + 24 = 0$

Grade 11 Quadratic Equations and Expressions CCSS: HSG-GPE.A.2

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Now, using

$(h+p,k)$ as the focus and

$x = h-p$ as the directrix, what would be the distance,

$d_1$ from the focus to the point

$(x,y)$ on the parabola?

$d_1 = sqrt( (x-h)^2 + (y-k)^2 )$
$d_1 = sqrt( (x-(h+p))^2 + (y-k)^2 )$
$d_1 = sqrt( (x-(h+p)^2))$
$d_1 = sqrt( (y-k)^2 )$

Grade 11 Quadratic Equations and Expressions CCSS: HSG-GPE.A.2

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What would be the distance,

$d_2$ , from the point

$(x,y)$ to the directrix

$x=h-p ?$

$d_2 = sqrt((x-(h-p))^2)$
$d_2 = sqrt( (x-h)^2 )$
$d_2 = sqrt(x^2 + y^2)$
$d_2 = sqrt( (x-(h-p))^2 + (y-k)^2)$

Grade 11 Quadratic Equations and Expressions CCSS: HSG-GPE.A.2

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Using the equation found in the fourth question, what is the resulting equation once

$d_1, d_2$ are substituted for the equations found for them in the previous two questions? Is this the final equation for a parabola with focus

$(h+p, k)$ and directrix

$x=h-p$ , or is more work needed?

$(y-k)^2 = 4p(x+h) \ \$ This is the final equation.
$(y-k)^2 = 2h(p-2x) \ \$ This is the final equation, but looks different than normal because the focus was $(h+p,k)$ instead of $(h,k)$ .
$(y-k)^2 = 4p(x-h) \ \$ This is the final equation.
$(y-k) = 4p(x-h)^2 \ \$ This is not the final equation, but it simply needs to be fully expanded and rearranged.

Grade 11 Quadratic Equations and Expressions CCSS: HSG-GPE.A.2

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What is the reason in the seventh question for the focus being

$(h+p,k)$ and the directrix being

$x=h-p$ , instead of just being arbitrary constants? Would the derivation still have been possible using arbitrary constants?

Using the fact that the focus and directrix are equal distances, p, from the vertex, (h, k), this results in the more common version of the equation for a parabola. The derivation would still have been possible, but the resulting form would not have looked familiar.
Knowing that the common final form of the equation has the letter p in it, this is included in the focus and directrix, adding in one and subtracting in the other so that they cancel out. The derivation cannot be done without this alteration.
Since the axis of symmetry of the parabola is perpendicular to the directrix, we use this fact to come up with the p value used in these definitions of the focus and directrix. The derivation would have been identical without this, but is usually included as an interesting mathematical fact.
There is no reason for this, these are completely arbitrary.

Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.2

What is the equation of a parabola with the focus at

$(1,3)$ and a directrix of

$x=-5?$

$(x-3)^2 = 12(y+2)$
$(y-3)^2 = 12(x-1)$
$(y+3)^2 = 3(x-2)$
$(y-3)^2 = 12(x+2)$

Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.2

What is the equation of a parabola with the focus at

$(4,-5)$ and directrix

$x=0?$

$(y-5)^2 = 8x$
$(y+5)^2 = 8(x - 4)$
$(y+5)^2 = 8x - 16$
$(y-5)^2 + 8 = 4x$

Grade 11 Quadratic Equations and Expressions CCSS: HSG-GPE.A.2

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Which of the following is the best definition of a parabola?

A U-shaped curve.
For a given point and line not through the point, the set of all points equidistant from this given point and line.
The object contained by a focal point and directrix when shown in a graphic form.
An equation with an $x^2$ term.