Discuss the equation
2x^2-y^2+4x+4y-4=0
centrr, transverse axis, vertices foci TWo point s equation of asymptope
x=2cost
y=3cost
0 is less than or equal to t and t is less than or equal to pi. Find rectangular equation
2x^2-y^2+4x+4y-4=0
2x^2+4x - y^2+4y - 4 = 0
2x^2+4x+2 - y^2+4y-4 -4-2+4 = 0
2(x+1)^2 - (y-2)^2 = 2
(x+1)^2 - (y-2)^2/2 = 1
Now just consult the properties of hyperbolas.
x = 2cost
y = 3cost
First off, I think you have a typo, since that is just 3x=2y
If you meant
x = 2cost
y = 3sint, then you have
x/2 = cost
y/3 = sint
and we all know that cos^2t + sin^2t = 1
To begin, let's analyze the given equation:
2x^2 - y^2 + 4x + 4y - 4 = 0
To identify the type of conic section represented by this equation, we can check the coefficients of x^2 and y^2. In this case, since the x^2 coefficient is positive and the y^2 coefficient is negative, we can conclude that it is an ellipse.
To rewrite the equation in a standardized form, we need to complete the square separately for the x and y terms. Let's start with the x terms:
2x^2 + 4x = 0
Divide both sides of the equation by 2:
x^2 + 2x = 0
Now, add the square of half the coefficient of x, which is (2/2)^2 = 1, to both sides:
x^2 + 2x + 1 = 1
Simplify:
(x + 1)^2 = 1
Next, let's complete the square for the y terms:
-y^2 + 4y = 4
Multiply both sides of the equation by -1 to make the leading coefficient positive:
y^2 - 4y = -4
Add the square of half the coefficient of y, which is (4/2)^2 = 4, to both sides:
y^2 - 4y + 4 = 0
Simplify:
(y - 2)^2 = 0
Now, let's rewrite the entire equation in standardized form:
(x + 1)^2 = 1
(y - 2)^2 = 0
From these equations, we can identify the center of the ellipse as (-1, 2). Notice that the equation for y has a squared term equal to zero, which indicates that the ellipse is a "degenerate" ellipse. In this case, it reduces to a single point on the y-axis at (x=-1, y=2).
To find the transverse axis, vertices, and foci, we can use the general form of an ellipse equation:
[(x - h)^2 / a^2] + [(y - k)^2 / b^2] = 1
Where (h, k) represents the center of the ellipse, a represents the semi-major axis, and b represents the semi-minor axis.
Comparing this with the given equation, we can determine that the semi-major axis (a) is 1 and the semi-minor axis (b) is 0. Since the semi-minor axis is zero, this also indicates a degenerate ellipse.
Therefore, the center of the ellipse is (-1, 2), the transverse axis is along the x-axis, the vertices are at (-1±1, 2), and the foci coincide with the center.
Next, you mentioned the parametric equations:
x = 2cos(t)
y = 3cos(t)
These parametric equations describe an ellipse centered at the origin, with semi-major axis 2 and semi-minor axis 3.
Finally, you asked for the rectangular equation given these parametric equations. To find this, we can eliminate the parameter t by rearranging the equations:
From x = 2cos(t), we can solve for cos(t) as cos(t) = x/2.
Similarly, from y = 3cos(t), we can solve for cos(t) as cos(t) = y/3.
Setting these two equivalent expressions for cos(t) equal to each other, we have:
x/2 = y/3
Now, we can rearrange this equation to get the rectangular equation:
3x = 2y
So, the rectangular equation for the given parametric equations is 3x - 2y = 0.
I hope this explanation helps! Let me know if you have any further questions.