What conic section is drawn by the parametric equations x=csc t and y cot?
A. Parabola
B. Circle
C. Ellipse
D. Hyperbola
If your expression mean:
x = csc t and y = cot t
then:
x = csc t
x = 1 / sin t
sin t = 1 / x
y = cot t = cos t / sin t = ± √ ( 1 - sin² t ) / sin t = ± √ ( 1 - sin² t ) ∙ 1 / sin t =
± √ ( 1 - sin² t ) ∙ csc t = ± √ [ 1 - ( 1 / x )² ] ∙ x = ± √ ( x² / x² - 1 / x² ) ∙ x =
± √ [ ( x² - 1 ) / x² ] ∙ x =± [ √ ( x² - 1 ) / √ x² ] ∙ x =
± [ √ ( x² - 1 ) / x ] ∙ x = ± √ ( x² - 1 )
y = ± √ ( x² - 1 )
hyperbola
To determine the conic section drawn by the given parametric equations x=csc t and y=cot t, we need to eliminate the parameter t.
Starting with x = csc t, we can rewrite it as sin t = 1/x.
Next, y = cot t can be expressed as y = cos t / sin t.
Substituting sin t = 1/x into y = cos t / sin t yields y = cos t / (1/x) = x cos t.
This equation is equivalent to y = x cot t.
Now we have y = x cot t, which can also be written as yx = cot t.
Rationalizing the denominator gives yx = cos t / sin t.
Since sin t = 1/x, it becomes yx = cos t / (1/x) = x cos t.
Rearranging, we get yx = x cos t, which simplifies to y = cos t.
Therefore, the conic section drawn by the given parametric equations is a line.
The correct answer is: None of the above.
To determine the conic section drawn by the parametric equations x = csc(t) and y = cot(t), we can start by eliminating the parameter t and obtaining an equation in terms of x and y.
Let's start with the equation x = csc(t). Since csc(t) is the reciprocal of sine, we can rewrite it as x = 1/sin(t).
Similarly, for the equation y = cot(t), we know that cot(t) is the reciprocal of tangent, so we can rewrite it as y = 1/tan(t).
To eliminate t, we can take the reciprocal of both x and y equations.
Reciprocal of x = 1/(1/sin(t)) = sin(t).
Reciprocal of y = 1/(1/tan(t)) = tan(t).
So, we have sin(t) = x and tan(t) = y as the resulting equations.
Now, we can square both equations to eliminate the trigonometric functions:
(sin(t))^2 = x^2
(tan(t))^2 = y^2
We know that sin^2(t) + cos^2(t) = 1, and dividing both sides of this equation by cos^2(t), we get:
sin^2(t)/cos^2(t) + cos^2(t)/cos^2(t) = 1/cos^2(t)
tan^2(t) + 1 = sec^2(t)
Rearranging this equation, we have:
1 + tan^2(t) = sec^2(t)
Comparing this with our equations (tan(t))^2 = y^2 and (sin(t))^2 = x^2, we can see that this indicates a hyperbola.
Therefore, the conic section drawn by the parametric equations x = csc(t) and y = cot(t) is a hyperbola.
D. Hyperbola.