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2 years ago

Which is the equation of a hyperbola with directrices at x = ±3 and foci at (4, 0) and (−4, 0)?

Mathematics

1 answer:

tester [92]2 years ago

8 0

Answer:

x^2/12 - y^2/4 = 1

Step-by-step explanation:

As the diretrices have simetrical values of x and have y = 0, the center is located at (0,0)

The formula for the diretrices is:

x1 = -a/e and x2 = a/e

And the foci is located at (a*e, 0) and (-a*e, 0)

So we have that:

a/e = 3

a*e = 4

From the first equation, we have a = 3e. Using this in the second equation, we have:

3e*e = 4

e^2 = 4/3

e = 1.1547

Now finding the value of a, we have:

a = 3*1.1547 = 3.4641

Now, as we have that b^2 = a^2*(e^2 - 1), we can find the value of b:

b^2 = 3.4641^2 * (1.1547^2 - 1) = 4

b = 2

So the equation of the hyperbola (with vertical diretrices and center in (0,0)) is:

x^2/a^2 - y^2/b^2 = 1

x^2/12 - y^2/4 = 1

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Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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Annie is creating a stencil for her artwork using a coordinate plane. The beginning of the left edge of the stencil falls at (2,

taurus [48]

Answer:

(A)(12, 9)

Step-by-step explanation:

Given:

The beginning of the left edge of the stencil falls at (2, −1).

A point, say Q on the stencil is at (4, 1).

Point Q divides the stencil into the ratio 1:4.

We are required to find the end of the stencil.

Mathematically, Point Q divides the stencil internally in the ratio 1:4.

For internal division of a line with beginning point $(x_1,y_1)$ and end point $(x_2,y_2)$ in the ratio m:n, we use the formula

$Q(x,y)=(\dfrac{mx_2+nx_1}{m+n} ,\dfrac{my_2+ny_1}{m+n} )$

$(x_1,y_1)=(2, -1)$ , $(x_2,y_2)=?$ , Q(x,y)=(4,1), m:n=1:4

Therefore:

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The correct option is A.

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