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3 years ago

Which equations represent the asymptotes of the hyperbola?

Mathematics

1 answer:

makvit [3.9K]3 years ago

4 0

Answer:

see below

Step-by-step explanation:

The equation of the hyperbola can be written as ...

((x -h)/a)² -((y -k)/b)² = 1

This has asymptotes ...

(x -h)/a ± (y -k)/b = 0

Solving for y, we have ...

y = ±(b/a)(x -h) +k

Filling in the given values a=6, b=8, h=1, k=2, we have ...

y = ±8/6(x -1) +2

$y=\dfrac{\pm4x\mp4+6}{3}\\\\\boxed{y=\dfrac{4x+2}{3}\ \text{and }y=\dfrac{10-4x}{3}}$

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Step-by-step explanation:

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2 years ago

Read 2 more answers

I don’t really get this one.<br> Thx and have a great day!!

gayaneshka [121]

Answer:

The correct option is C). (9,4)

The coordinates of a point N is (9,4)

Step-by-step explanation:

Theory: If point P(x,y) lies on line segment AB and AP: PB=m:n, then we say P divides line AB internally in ratio of m:n and Point is given by

P= $(\frac{mX2+nX1}{m+n} , \frac{mY2+nY1}{m+n})$

Given that point, M is lying somewhere between point L and point N.

The coordinates of a point L is (-6,14)

The coordinates of a point M is (-3,12)

Also, LM: MN = 1:4

We can write as,

Let,

Point L(-6,14)=(X1, Y1)

Point M(-3,12)=(x,y)

Point N is (X2, Y2)

m=1 and n=4

M(-3,12)= $(\frac{mX2+nX1}{m+n} , \frac{mY2+nY1}{m+n})$

M(-3,12)= $(\frac{1(X2)+4(-6)}{1+4} , \frac{(Y2)+4(14)}{1+4})$

M(-3,12)= $(\frac{(X2)-24}{5} , \frac{1(Y2)+56}{5})$

$(-3)=\frac{(X2)-24}{5}$

(-15)=X2-24

X2=9

$(12)=\frac{(Y2)+56}{5}$

(60)=Y2+56

Y2=4

Thus,

The coordinates of a point N is (9,4)

Result: The correct option is C). (9,4)

3 0

3 years ago