Identify the type of conic section that has the equation 9x^2+ 25y^2 =225 and identify its domain and range.
9x^2+ 25y^2 =225
divide each term by 225
x^2/25 + y^2/9 = 1
ellipse, with a = 5 and b = 3
domain: -5 ≤ x ≤ 5
range: -3 ≤ y ≤ 3
Ah, the good ol' conic section question! The equation you have there, 9x^2 + 25y^2 = 225, represents an ellipse. It's like the rounder and friendlier cousin of the hyperbola.
As for its domain and range, since x^2 and y^2 are both positive, the values of x and y can be anything from negative infinity to positive infinity. So, the domain and range for this ellipse are, drumroll please... all real numbers! So, go ahead and let your imagination run wild within the endless realm of this elliptical creature. Enjoy!
The equation 9x^2 + 25y^2 = 225 represents an ellipse.
To identify its domain and range, let's first solve for y^2:
25y^2 = 225 - 9x^2
Dividing both sides by 25, we have:
y^2 = (225 - 9x^2) / 25
Taking the square root of both sides, we get:
y = ± √[(225 - 9x^2) / 25]
The domain of the ellipse is the values that x can take. To find this, we need to determine the interval of x values that satisfy the equation.
First, solve for x^2:
9x^2 = 225 - 25y^2
Dividing both sides by 9, we have:
x^2 = (225 - 25y^2) / 9
Taking the square root of both sides, we get:
x = ± √[(225 - 25y^2) / 9]
The range of the ellipse is the values that y can take. To determine this, we'll consider the expression inside the square root, (225 - 9x^2) / 25.
Since the equation implies that y^2 has to be positive, we can deduce that the range is y ∈ (-∞, ∞).
Therefore, the domain of the ellipse is x ∈ (-√(225/9), √(225/9)), and the range is y ∈ (-∞, ∞).
To identify the type of conic section represented by the equation 9x^2 + 25y^2 = 225, we can analyze the equation and its coefficients.
First, let's observe the coefficients of x^2 and y^2:
- The coefficient of x^2 is positive (9 > 0)
- The coefficient of y^2 is also positive (25 > 0)
Since both coefficients are positive and equal, we can conclude that the conic section represented by this equation is an ellipse.
Now, let's determine the domain and range of the ellipse.
The equation 9x^2 + 25y^2 = 225 can be rearranged as:
x^2/25 + y^2/9 = 1
Comparing it with the standard form of the ellipse equation, (x - h)^2/a^2 + (y - k)^2/b^2 = 1, we can deduce that:
- The center of the ellipse is (h, k) = (0, 0)
- The semi-major axis length a = √25 = 5 (corresponds to the horizontal axis)
- The semi-minor axis length b = √9 = 3 (corresponds to the vertical axis)
From this information, we can determine:
- The domain: The values of x that satisfy the equation. Since x^2/25 ≤ 1, the domain of this ellipse is [-5, 5].
- The range: The values of y that satisfy the equation. Since y^2/9 ≤ 1, the range of this ellipse is [-3, 3].
Thus, the domain of the ellipse is [-5, 5], and the range is [-3, 3].