Find an equation in standard form for the hyperbola with vertices at (0, ±9) and foci at (0, ±10).
1 answer:
Answer:
y² / 81 - x² / 19 = 1
Step-by-step explanation: See Annex ( vertices and foci in coordinates axis)
The equation in standard form for the hyperbola is:
x² / a² - y²/b² = 1 or y²/a² - x² / b² = 1
In cases of transverse axis parallel to x axis or y axis respectively.
As per given information in this case hyperbola has a transverse axis parallel to y axis the equation is
y²/a² - x² / b² = 1
a is a distance between center and vertex therefore a = 9
c is a distance between center and a focus c = 10
and b will be:
c² = a² + b² ⇒ b² = c² - a² ⇒ b² = (10)² - (9)² ⇒ b² = 100 - 81
b = √19
And the equation in standard form is:
y² / a² - x² / b² = 1
y² / ( 9 )² - x² / √(19)² ⇒ y² / 81 - x² / 19 = 1
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Answer:
Option D
Step-by-step explanation:
f(x) =
Transformed form of the function 'f' is 'g'.
g(x) =
Property of vertical stretch or compression of a function,
k(x) = x
Transformed function → m(x) = kx
Here, k = scale factor
1). If k > 1, function is vertically stretched.
2). If 0 < k < 1, function is vertically compressed.
From the given functions, k =
Since, k is between 0 and , function f(x) is vertically compressed by a scale factor
.
g(x) = f(x + 4) represents a shift of function 'f' by 4 units left.
g(x) = f(x - 4) represents a shift of function 'f' by 4 units right.
g(x) =
Therefore, function f(x) has been shifted by 4 units left to form image function g(x).
Option D is the answer.