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Anna [14]
3 years ago
7

A radio telescope has a parabolic surface, as shown below.If the telescope is 1 m deep and 8 m wide, how far is the focus from t

he vertex?

Mathematics
1 answer:
scoray [572]3 years ago
3 0

Answer:

  4 m

Step-by-step explanation:

The equation can be written ...

  y = (x/4)² = (1/16)x²

where the divisor 4 is the horizontal scale factor that makes the parabola 8 m wide at y = 1 m.

This can also be written as ...

  y = 1/(4p)x² = (1/16)x²

from which we can see ...

  4p = 16

  p = 4

In this form, p is the distance from the vertex to the focus, 4 meters.

_____

Another way to find the y-coordinate of the focus is to draw a line with slope 1/2 through the vertex. It will intersect the parabola at the point where the vertical distance to the directrix is the same as the horizontal distance to the focus. The y-coordinate of that point is the y-coordinate of the focus.

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The railroad company wants to sell $8,134 in tickets. If each ticket costs $7, how many
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1162 tickets will need to be sold

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8134 / 7 = 1,162

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Square ABCD has vertices at A(-8,1), B(3,6), and D(-3,-10). What are the coordinates of point c?
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3 years ago
X=y^2 is x a function of y
oksano4ka [1.4K]

It is false that x is a function of y

<h3>How to determine the true statement?</h3>

The equation is given as:

x = y^2

Set y = 2

x = (2)^2

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Set y = -2

x = (-2)^2

x = 4

Different values of y give the same x value.

This means that x is not a function of y

Read more about functions at:

brainly.com/question/12431044

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3 0
2 years ago
12)If p and p' are the length of the perpendiculars drawn from the points (+-root a2-b2,0) to the line x/a cosA + y/b sinA = 1,
Elden [556K]

Answer:

Kindly refer to below explanation.

Step-by-step explanation:

Given equation of line:

\dfrac{x}{a}cos\theta+ \dfrac{y}{b}sin\theta =1\\OR\\\dfrac{x}{a}cos\theta+ \dfrac{y}{b}sin\theta-1=0

Given points are:

(\sqrt{a^2-b^2},0), (-\sqrt{a^2-b^2},0)

Formula for distance between a point and a line is:

If the point is (m, n) and equation of line is Ax+By+C = 0

Then, perpendicular distance between line and points is:

Distance = \dfrac{|Am+Bn+C|}{\sqrt{A^2+B^2}}

Here,

A = \dfrac{x}{a}cos\theta\\B = \dfrac{y}{b}sin\theta\\C = -1

For the first point:

m = \sqrt{a^2-b^2}, n = 0

By the above formula:

p and p' can be calculated as:

p\times p' =\\\Rightarrow  \dfrac{|\dfrac{cos\theta}{a}\times \sqrt{a^2-b^2}+ 0 -1|}{\sqrt{\dfrac{cos^2\theta}{a^2}+\dfrac{sin^2\theta}{b^2}}}\times \dfrac{|-\dfrac{cos\theta}{a}\times \sqrt{a^2-b^2}+ 0 -1|}{\sqrt{\dfrac{cos^2\theta}{a^2}+\dfrac{sin^2\theta}{b^2}}}\\\Rightarrow \dfrac{1-(\sqrt{a^2-b^2}\times \frac{cos\theta}{a})^2}{\dfrac{cos^2\theta}{a^2}+\dfrac{sin^2\theta}{b^2}}}

Formula used:

(x+y)(x-y) = x^2 - y^2

\Rightarrow \dfrac{\dfrac{a^2-a^2cos^2\theta+b^2cos^2\theta}{a^2}}{\dfrac{b^2cos^2\theta+a^2sin^2\theta}{a^2b^2}}\\\Rightarrow b^2(\dfrac{a^2sin^2\theta+b^2cos^2\theta}{a^2sin^2\theta+b^2cos^2\theta})\\\Rightarrow b^2

(Using the identity:

sin^2\theta +cos^2\theta = 1)

(Hence provded)

5 0
2 years ago
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