3 years ago

What is the measurement of FKH?

Mathematics

1 answer:

stepan [7]3 years ago

6 0

Answer:

Step-by-step explanation:

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16: Find the coordinates of the foci and the vertices, the eccentricity, the length of the latus rectum of the Hyperbola y^2-16x

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Answer:

Foci $(0,\pm\frac{\sqrt{17}}{4})$

Vertices: $(0,\pm 1)$

Eccentricity, e= $\frac{\sqrt{17}}{4}$

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Step-by-step explanation:

We are given that

$y^2-16x^2=1$

$y^2-\frac{x^2}{(\frac{1}{4})^2}=1$

The equation of hyperbola is along y-axis because y is positive

Compare the equation with

$y^2/a^2-x^2/b^2=1$ (Along y-axis)

We get

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$\frac{16+1}{16}=c^2$

$c^2=\frac{17}{16}$

$c=\pm \frac{\sqrt{17}}{4}$

Therefore,

The coordinates of foci= $(0,\pm c)=(0,\pm\frac{\sqrt{17}}{4})$

The coordinated of vertices= $(0,\pm a)=(0,\pm 1)$

Eccentricity, e=c/a

$e=\frac{\frac{\sqrt{17}}{4}}{1}=\frac{\sqrt{17}}{4}$

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