Answer:
Step-by-step explanation:
Solve for :
-Multiply both sides of the equation by the denominator :
-Subtract both sides by :
So, the value of is
.
What type of triangle can a right triangle be?
1 answer:
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So... if you notice the picture below, based on the given vertices, is a hyperbola with a vertical traverse axis
meaning for the equation, the fraction with the "y" variable is the positive fraction, thus
so... what' the point h,k for the center? well, just take a peek at the graph, the center is half-way between both vertices, that's h,k
what's the "a" component or the traverse axis... well, is the distance from one vertex to the center, notice the picture, notice how many units from either vertex to the center, that's "a"
now.. what's "b"? well, "b" comes from the conjugate axis, or the other runnning over the x-axis hmm the smaller one in this case.... well, we dunno what "b" is
however, we know the distance from either focus, to the center of the hyperbola, and that distance "c", is
notice the picture, notice the distance from either focus to the center, that's the distance "c", thus
that'd be "b"
bearing in mind that
meaning for the equation, the fraction with the "y" variable is the positive fraction, thus
so... what' the point h,k for the center? well, just take a peek at the graph, the center is half-way between both vertices, that's h,k
what's the "a" component or the traverse axis... well, is the distance from one vertex to the center, notice the picture, notice how many units from either vertex to the center, that's "a"
now.. what's "b"? well, "b" comes from the conjugate axis, or the other runnning over the x-axis hmm the smaller one in this case.... well, we dunno what "b" is
however, we know the distance from either focus, to the center of the hyperbola, and that distance "c", is
notice the picture, notice the distance from either focus to the center, that's the distance "c", thus
that'd be "b"
bearing in mind that