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Radda [10]
3 years ago
5

Using a directrix of y = −3 and a focus of (2, 1), what quadratic function is created?

Mathematics
2 answers:
posledela3 years ago
8 0

Answer:

f(x) = one eigth (x-2)^2 - 1

Step-by-step explanation:

We have. focus = (2,1) and directrix y = -3.

Let (x,y) be any point on the curve.

First, we will find the distance between (x,y) and the focus (2,1).

i.e. distance = \sqrt{(x-2)^{2} +(y-1)^{2} }

Now, we will find the distance between (x,y) and directrix y = -3.

i.e. distance = | y - (-3) | = | y + 3 |

To find the quadratic function, we will equate both the distances and simplify the equation.

i.e. \sqrt{(x-2)^{2} +(y-1)^{2} } = | y + 3 |

Squaring both sides, we get,

(x-2)^{2}+(y-1)^{2}=(y+3)^{2}

i.e. x^{2}-4x-8y-4=0

i.e. 8y=x^{2}-4x-4

i.e. y=\frac{x^{2} }{8} -\frac{x}{2} -\frac{1}{2}

The option f(x) = one eigth (x-2)^2 - 1 will give the y obtained above after simplifying.

bazaltina [42]3 years ago
3 0
<span>f(x) = one eighth (x - 2)^2 - 1
   Since a parabola is the curve such that all points on the curve have the same distance from the directrix as the distance from the point to the focus.With that in mind, we can quickly determine 3 points on the parabola. The 1st point will be midway between the focus and the directrix, So: (2, (1 + -3)/2) = (2, -2/2) = (2,-1). The other 2 points will have the same y-coordinate as the focus, but let offset on the x-axis by the distance from the focus to the directrix. Since the distance is (1 - -3) = 4, that means the other 2 points will be (2 - 4, 1) and (2 + 4, 1) which are (-2, 1) and (6, 1). The closest point to the focus will have the same x-coordinate as the focus, so the term will be (x-2)^2. This eliminates the functions "f(x) = -one eighth (x + 2)^2 - 1" and "f(x) = -one half (x + 2)^2 - 1" from consideration since their x term is incorrect, leaving only "f(x) = one eighth (x - 2)^2 - 1" and "f(x) = one half (x - 2)^2 + 1" as possible choices. Let's plug in the value 6 for x and see what y value we get from squaring (x-2)^2. So: (x-2)^2 (6-2)^2 = 4^2 = 16 Now which option is equal to 1? Is it one eighth of 16 minus 1, or one half of 16 plus 1? 16/8 - 1 = 2 - 1 = 1 16/2 + 1 = 8 + 1 = 9 Therefore the answer is "f(x) = one eighth (x - 2)^2 - 1"</span>
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Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\

With respect to reference angle A, we have:

  • opposite side = a = 4
  • adjacent side = b = \sqrt{33}
  • hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\

Then cosine which is adjacent over hypotenuse

\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\

Tangent is the ratio of opposite over adjacent

\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

  • cosecant, abbreviated as csc, is the reciprocal of sine
  • secant, abbreviated as sec, is the reciprocal of cosine
  • cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{  ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\

------------------------------------------------------

Summary:

The missing side is b = \sqrt{33}

The 6 trig functions have these results

\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

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1 year ago
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