If the equation of the hyperbola is (9x^2)-(64y^2) = 76., the point P (-8, 0 ) is its ?

Question

Yes! Northsta.. You are quite right :
the correct number is the one indicated by you and not 76 ! Spelling error !
Apologies !
Also your answer is correct : vertex !
Thanks
Bye

Answer 1

Are you sure the equation should be equal to 76 instead of 576?

9x² - 64y² = 76

Divide by 576.

x²/64 - y²/9 = 76/576

If the equation is correct as written, the point (-8,0) is not on the curve and nothing special. If the equation should be equal to 576 then the point (-8,0) is one of the vertices of the pair of hyperbolas.

Answer 2

(9x^2)-(64y^2) = 76
we want equation into form;
x^2/a^2-y^2/b^2=1
a^2=76/9,b^2=76/64
b/a=+/-3/8
the equation of the
asymptote
y=+/-3/8x
these lines touch the curve
of the hyperbola at four
points at x= -infinity and +infinity
the eccentricity
e=sqrt(b^2/a^2+1)
=sqrt(35)/4
foci S and S'=+/-a*e
=+/-sqrt(76*35)/12
directrix D and D'
=+/-a/e
=+/-2*sqrt(76/35)

sorry, i can't find the point
P(+/-8,0) that has any
significant meaning in this
hyperbola
[if the equation is
9x^2+64y^2=576 as suggested
by northstar,substituting
y=0 gives x=+/-8{a=8,b=3}
-we then have the points
P(-8,0)andQ(8,0) which are the
vertices of the west and east
hyperbolas-that is,the points
where the curves cross the
x-axis {or the transverse axis
of the hyperbola}]

i hope that this helps

Answer 3

10 past 4

Answer 4

that doesn't make sense. the point P is nothing - its certainly not on the hyperbola. more details?

Answer 5

I suspect that it is one focus but I haven't worked through to check.

Answer 6

well if you multiply that by E=MCsquared and take away
the number you first thought,you should get the answer.

Answer 7

centre?