Find an asymptote of this conic section. 9x^2-36x-4y^2+24y-36=0

1 answer:

8 0

We will begin by grouping the x terms together and the y terms together so we can complete the square and see what we're looking at. $(9x^2-36x)-(4y^2+24y)-36=0$ . Now we need to move that 36 over by adding to isolate the x and y terms. $(9x^2-36x)-(4y^2+24y)=36$ . Now we need to complete the square on the x terms and the y terms. Can't do that, though, til the leading coefficients on the squared terms are 1's. Right now they are 9 and 4. Factor them out: $9(x^2-4x)-4(y^2-6y)=36$ . Now let's complete the square on the x's. Our linear term is 4. Half of 4 is 2, and 2 squared is 4, so add it into the parenthesis. BUT don't forget about the 9 hanging around out front there that refuses to be forgotten. It is a multiplier. So we are really adding in is 9*4 which is 36. Half the linear term on the y's is 3. 3 squared is 9, but again, what we are really adding in is -4*9 which is -36. Putting that altogether looks like this thus far: $9(x^2-4x+4)-4(y^2-6y+9)=36+36-36$ . The right side simplifies of course to just 36. Since we have a minus sign between those x and y terms, this is a hyperbola. The hyperbola has to be set to equal 1. So we divide by 36. At the same time we will form the perfect square binomials we created for this very purpose on the left: $\frac{(x-2)^2}{4}- \frac{(y-3)^2}{9}=1$ . Since the 9 is the bigger of the 2 values there, and it is under the y terms, our hyperbola has a horizontal transverse axis. a^2=4 so a=2; b^2=9 so b=3. Our asymptotes have the formula for the slope of $m=+/- \frac{b}{a}$ which for us is a slope of negative and positive 3/2. Using the slope and the fact that we now know the center of the hyperbola to be (2, 3), we can solve for b and rewrite the equations of the asymptotes. $3= \frac{3}{2}(2)+b$ give us a b of 0 so that equation is y = 3/2x. For the negative slope, we have $3=- \frac{3}{2}(2)+b$ which gives us a b value of 6. That equation then is y = -3/2x + 6. And there you go!

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In a circle with a radius of 12.6 ft, an arc is intercepted by a central angle of 2 radians

Gennadij [26K]

Question:

In a circle with a radius of 12.6 ft, an arc is intercepted by a central angle of 2π/7 radians.

What is the arc length?

Use 3.14 for π and round your final answer to the nearest hundredth.

Enter your answer as a decimal in the box.

Answer:

<h3>Arc length is 11.30 feet</h3>

Solution:

Given that,

Radius of circle = 12.6 feet

Central angle = $\frac{2 \pi }{7}$ radians

To find: Arc length

The arc length of a circle of radius "r" when central angle given in radians is:

$s = r \times \theta$

Where,

s is the arc length

r is the radius

$\theta$ is the central angle in radians

Substituting the values we get,

$s = 12.6 \times \frac{2 \times \pi }{7}\\\\s = 12.6 \times \frac{2 \times 3.14 }{7}\\\\x = 11.304 \approx 11.30$

Thus, arc length is 11.30 feet

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3 years ago

A circle is defined by the equation x2 + y2 – x – 2y – = 0. What are the coordinates for the center of the circle and the length

madreJ [45]

X^2 + y^2 - x - 2*y = 0
To find both coordinates and radius we need to make this equation in circle form:
(x-a)^2 + (y-b)^2 = r^2

x^2 - 2*1/2*x + 1/4 - 1/4 + y^2 - 2*1*y + 1 - 1 = 0
Here we are adding and subtracting numbers in order to get square binomial.
(x - 1/2)^2 + (y-1)^2 = 5/4

coordinates of center are (1/2,1) and the radius is √ $\frac{5}{4}$