What are the foci of the hyperbola given by the equation 16x2 - 25y2 - 64x - 50y - 361 = 0?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

What are the foci of the hyperbola given by the equation 16x2 - 25y2 - 64x - 50y - 361 = 0?

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

complete the squares
How?
(-1 ± , 2) (2 ± , -1) (-1, 2 ± ) (2, -1 ± )

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Its one of these lol Im just not sure which
|dw:1355058848205:dw|
first factor the coeficientes of the squared terms (i re-arranged the terms) \[ \large 16x^2-64x-25y^2-50=361 \] \[ \large 16(x^2-4x+\qquad)-25(y^2+2y+\qquad)=361 \]
got this?
Kinda but what are the blank spaces?
there's a "y" missing next to the 50.
oh okay
Pretty sure its A or B right?
now in each parenthesis there's a blank. u have to write there coeficient of the linear term (no power) halved and then squared \[ \large \color{green}{16}(x^2-4x+\color{red}{4})\color{orange}{-25}(y^2+2y+\color{red}{1})=361 +\color{green}{16}\cdot\color{red}{4}\color{orange}{-25}\cdot\color{red}{1} \]
got this?
http://en.wikipedia.org/wiki/Hyperbola
mhm :)
now \[ \large 16(x-\color{red}{2})^2-25(y\color{red}{+1})^2=400 \] \[ \large \frac{16(x-\color{red}{2})^2}{400}-\frac{25(y-(\color{red}{-1}))^2}{400}=1 \] \[ \large \frac{(x-\color{red}{2})^2}{25}-\frac{(y-(\color{red}{-1}))^2}{16}=1 \]
got this?
I see :)
now the center of the hyperbola is \[ \large (\color{red}{2},\color{red}{-1}) \]
is it a horizontal or vertical hyperbola? how can u tell?
horizontal
the format of the equation.
no.
look at the denominators which is larger? under x? or under y?
x.....
yes. this tells u the hyperbola is horizontal
I said it was horizontal. Is it B?
what is the length of the semi-mayor axis?
it HAS to. do u know why?
because the formula is formatted (h±a,k) for horizontal and that fits? lol
yes. very clever. u were looking for the foci not the vertices, though.
did u understand?
do it is B? :) and yeah pretty much lol
so*
i hope so.

Not the answer you are looking for?

Search for more explanations.

Ask your own question