The equation of the hyperbola with directrices at x = ±2 and foci at (5, 0) and (−5, 0) is 
<h3>How to determine the equation of the hyperbola?</h3>
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
Read more about hyperbola at:
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Since triangle DEF = triangle JKL, m<D = m<J, m<E = m<K, m<F = m<L.
m<F = m<L = 90 degrees
m<K = m<E = 5(m<D)
but m<E + m<D = 90 degrees [right angled triangle]
5(m<D) + m<D = 90 degrees
6(m<D) = 90 degrees
m<D = 90 / 6 = 15 degrees.
Answer:
-12 = 3x - 4y
Step-by-step explanation:
Two points on the graph are (-4, 0) and (0, 3). Moving from the first point to the second, we see x (the 'run') increase by 4 and y (the 'rise') increase by 3. Thus, the slope of this line is m = rise / run = 3/4.
Using the point slope form, we get:
y - 3 = (3/4)(x - 0), or
4y - 12 = 3x, or
-12 = 3x - 4y (which is in Standard Form).
5) The relation between intensity and current appears linear for intensity of 300 or more (current = intensity/10). For intensity of 150, current is less than that linear relation would predict. This seems to support the notion that current will go to zero for zero intensity. Current might even be negative for zero intensity since the line through the points (300, 30) and (150, 10) will have a negative intercept (-10) when current is zero.
Usually, we expect no output from a power-translating device when there is no input, so we expect current = 0 when intensity = 0.
6) We have no reason to believe the linear relation will not continue to hold for values of intensity near those already shown. We expect the current to be 100 for in intensity of 1000.
8) Apparently, times were only measured for 1, 3, 6, 8, and 12 laps. The author of the graph did not want to extrapolate beyond the data collected--a reasonable choice.
Answer:
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