Find the equation of the hyperbola centered at the origin that satisfies the given conditions: x-intercepts = +,-4 and asymptote at y=3/4x

Question

Find the equation of the hyperbola centered at the origin that satisfies the given conditions: x-intercepts = +,-4 and asymptote at y=3/4x

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Mary 1 week 2021-10-07T00:55:42+00:00 2 Answers 0

Answers ( )

    0
    2021-10-07T00:57:05+00:00

    Answer:

    B

    Step-by-step explanation:

    edge

    0
    2021-10-07T00:57:07+00:00

    Answer:

    The equation of the hyperbola is presented as follows;

    \frac{x^2}{16} - \frac{y^2}{9} =1

    Step-by-step explanation:

    Here we have the standard equation of an hyperbola given as follows;

    \frac{x^2}{a^2} - \frac{y^2}{b^2} =1

    Where:

    a = x intercept

    The asymptote is ±(b/a)x

    Since the intercept, a is ± 4, the vertices are (-4, 0) and (4, 0)

    We are given the asymptote as y = 3/4x, therefore, since the genral form of the asymptote is ±(b/a)x, comparing, we have;

    ±(b/a)x ≡ 3/4x

    We have a = ±4, therefore, b = 3

    Hence the equation of the hyperbola is found by putting in the values of a and b in the general form as follows;

    \frac{x^2}{a^2} - \frac{y^2}{b^2} =1 \Rightarrow \frac{x^2}{4^2} - \frac{y^2}{3^2} =1

    The equation of the hyperbola = x²/16 – y²/9 = 1.

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