Question

Help

Answer 1

@Directrix

Answer 2

@phi

Answer 3

I really need help with this question

Answer 4

this one was painful. They don't say if they want )( or the other direction, so I'll assume sideways:

if the center is (4,1) the equation will be
\[ \frac{(x-4)^2}{a^2} - \frac{(y-1)^2}{b^2}= 1 \]

Answer 5

there is an equation for eccentricity
\[ \frac{\sqrt{a^2+b^2}}{a}= e = \frac{\sqrt{13}}{2}\]
square both sides
\[ \frac{a^2+b^2}{a^2}= \frac{13}{4} \]
multiply both sides by a^2
\[ a^2 +b^2 = \frac{13}{4} a^2\\b^2= \frac{9}{4}a^2
\]

Answer 6

ok

Answer 7

we want the curve to go through (8,4)
put that point into the equation
\[ \frac{(x-4)^2}{a^2} - \frac{(y-1)^2}{b^2}= 1 \\
\frac{16}{a^2} - \frac{9}{b^2} = 1\]
replace b^2 with 9a^2/4
\[ \frac{16}{a^2}- 9 \cdot \frac{4}{9a^2} = 1\]

almost there:
\[ \frac{16}{a^2}-\frac{4}{a^2} = 1 \\ \frac{12}{a^2}=1 \\ a^2=12\]

Answer 8

up above we have
\[ b^2= \frac{9}{4}a^2 \\ b^2= \frac{9}{4} \cdot 12 = 27\]

Answer 9

so the final equation is
\[ \frac{(x-4)^2}{12} - \frac{(y-1)^2}{27}= 1 \]

Answer 10

ok thanks

Answer 11

i was doing it different and keep getting the wrong answer

Answer 12

There might be a better way to do it. what way were you trying ?

Answer 13

so can we find back the value of "e" using that equation? you give ..

Answer 14

i was assuming a=2

Answer 15

since e=c/a

Answer 16

then a^2=4 so
c^2=a^2+b^2
13=4+b^2
9=b^2 like this

Answer 17

but with yours
a^2=12
b^2=27
then c^2=27+12
c^2=39
c=sq.rt 39

e=c/b
sq.rt39/sq.rt 27
im getting sq.rt 13/3 instead

Answer 18

ok but using e=c/a
we get sq.rt 39/sq.rt 12
=sq.rt 13/2 so urs must be correct

Answer 19

yes, I think c/a vs c/b might be which way the hyperbola is oriented
smile/frown versus ) ( shape

Answer 20

I did the same thing you did, but though we have the correct eccentricity, the curve does not go through point (8,4). So after some head scratching, I did it the way I posted up above.

Answer 21

so the major axis would be the x axis ?

Answer 22

The major axis goes through the vertexes, foci, and center.

Answer 23

would it be the y=1?

Answer 24

yes, y=1 is the axis of symmetry aka major axis