Please Check My Answers!!!
What is the type of conic section is given by the equation x^2-9y^2=900 and what is
the domain and range?
Answer: Type is hyperbola, Domain is all real values of x, not sure what the range is. Please explain how to find these
What are the vertex, focus and directrix of a parabola with equation: x = y2 + 14y − 2
the domain is clearly not all real numbers.
Suppose x=0. No way can -9y^2 = 900
The hyperbola is
x^2/900 - y^2/100 = 1
The vertices of the hyperbola are at (±30,0)
So, the domain is (-∞,-30]U[30,∞)
Using what you know of the shape of hyperbolas, the range is clearly all real numbers.
As for the parabola, recall the properties of the parabola
y^2 = 4px
Yours is shifted a bit, making it
(y+7)^2 = 1(x+51)
How did you get the second one?? Also, thank you so much for the first answer. I realized what I was doing wrong lol. I knew how to do it, I was just doing it wrong today
To determine the type of conic section given by the equation x^2-9y^2=900, you need to analyze the coefficients of x^2 and y^2. In this case, since the coefficient of x^2 is positive (1) and the coefficient of y^2 is negative (-9), it is a hyperbola.
Now, let's determine the domain and range of the equation. Since there are no restrictions on x in the equation x^2-9y^2=900, the domain is all real values of x.
To find the range, we can manipulate the equation to solve for y. Starting with x^2-9y^2=900, we can move the x^2 term to the other side and divide by -9, resulting in y^2 = (x^2 - 900) / -9. Taking the square root of both sides gives us y = ±sqrt((x^2 - 900) / -9).
From this equation, we can see that the range of y is dependent on the values of x. Since the square root of a negative number is undefined in real numbers, the range of y values will be all real numbers except for y values that make (x^2 - 900) / -9 negative. In this case, the range is y ∈ (-∞, ∞), meaning the equation covers all possible y values.
Moving on to the second question:
The given equation, x = y^2 + 14y - 2, represents a parabola.
To find the vertex, we need to transform the equation into vertex form. Completing the square on the equation will allow us to identify the vertex.
Starting with x = y^2 + 14y - 2, we can complete the square by adding (14/2)^2 = 7^2 = 49 to both sides of the equation:
x + 49 = y^2 + 14y + 49 - 2
Simplifying:
x + 49 = (y + 7)^2 + 47
Now, the equation is in vertex form, (y - k)^2 = 4a(x - h), where (h, k) represents the vertex. Comparing this form with our equation, we can see that the vertex is (-47, -7).
To find the focus and directrix of the parabola, we need to identify the value of a. In this case, a = 1/(4a) = 1/4.
Given that a = 1/4 and the vertex is (-47, -7), the focus is located at the point (-47, -7 + 1/4) = (-47, -27/4). The directrix is a horizontal line located at y = -7 - 1/4 = -29/4.
Therefore, the vertex of the parabola is (-47, -7), the focus is (-47, -27/4), and the directrix is y = -29/4.