Question #162061

 Draw the hyperbola with the equation: 3π‘₯ 2 βˆ’ 2𝑦 2 + 6π‘₯ βˆ’ 8𝑦 = 6


1
Expert's answer
2021-02-08T17:33:29-0500

first of all, we need to transform the equation to the canonical form

(xβˆ’x0)2a2βˆ’(yβˆ’y0)2b2=1\boxed{\frac {(x-x_0)^2} {a^2} - \frac{(y-y_0)^2} {b^2} = 1} , x0x_0 and y0y_0 are coordinates of the center,


3x2+6xβˆ’2y2βˆ’8y=63x^{2}+6x-2y^{2}-8y=6


(3x2+6x+3)βˆ’(2y2+8y+8)=1(3x^2 + 6x + 3) - (2y^2 +8y+ 8) = 1


(x+1)213βˆ’(y+2)212=1\dfrac{(x+1)^2}{\dfrac1 3} - \dfrac{(y+2)^2}{\dfrac 1 2} = 1


O (-1,-2) is the center

a=13β‰ˆ0.6a = \sqrt{\dfrac1 3} \thickapprox 0.6

b=12β‰ˆ0.7b = \sqrt{\dfrac1 2} \thickapprox 0.7


what is a and b:



(instead of the origin, there can be any point that is the center of the hyperbola)





Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Assignment Expert
09.02.21, 00:31

Dear JMD, we try to answer the question as soon as possible. You also can try to submit an order if special requirements should be met.

JMD
08.02.21, 14:09

hi, it would really make my day if someone was able to answer my question ASAP. Thanks

LATEST TUTORIALS
APPROVED BY CLIENTS