First, open the matrix calculator in the tool bar. Create this 1 x 3 matrix: 77 39 −33 Enter the value: Then select the operatio
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The attached image represents the triangle after it has been enlarged
<h3>How to enlarge the triangle?</h3>The given parameters are:
- Scale factor, k = -1/3
- Center, (a,b) = (-1,2)
From the graph, the coordinates of the triangle are:
A = (-4,2)
B = (-4,-4)
C = (-1,-4)
The rule of dilation is calculated using:
So, we have:
A' = (0, 2)
B' = (0, 4)
C' = (-1, 4)
See attachment for the image of the triangle, after it has been enlarged
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The equation of the hyperbola with directrices at x = ±2 and foci at (5, 0) and (−5, 0) is
The given parameters are:
- Directrices at x = ±2
- Foci at (5, 0) and (−5, 0)
The foci of a hyperbola are represented as:
Foci = (k ± c, h)
The center is:
Center = (h,k)
And the directrix is:
Directrix, x = h ± a²/c
By comparison, we have:
k ± c = ±5
h = 0
h ± a²/c = ±2
Substitute h = 0 in h ± a²/c = ±2
0 ± a²/c = ±2
This gives
a²/c = 2
Multiply both sides by c
a² = 2c
k ± c = ±5 means that:
k ± c = 0 ± 5
By comparison, we have:
k = 0 and c = 5
Substitute c = 5 in a² = 2c
a² = 2 * 5
a² = 10
Next, we calculate b using:
b² = c² - a²
This gives
b² = 5² - 10
Evaluate
b² = 15
The hyperbola is represented as:
So, we have:
Evaluate
Hence, the equation of the hyperbola is
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