Question

4x^2-5y^2-16x-30y-9=0
What is the standards form of the equation of the conic given above?

Answer 1

help!(:

Answer 2

anyone?

Answer 3

first, do you know the type of conic this equation represents?

Answer 4

no

Answer 5

this equation is in general form. we need to get it to standard form by completing the square. group all terms with x's together and y's together with parenthesis. what you got?

Answer 6

do we factor them to when group together?

Answer 7

that's later... just group them according to what i said.... also, move the constant over to the right side of the equal sign...

Answer 8

so we have..
(4x^2-16x)-(5y^2+30y)=-9

Answer 9

my bad..positive 9

Answer 10

ok.... now factor out a 4 from the x group then factor out a 5 from the y group. what u got now?

Answer 11

4x(x^2-4)-5y(y-6)=9

Answer 12

no... just a 4 and just a 5....

Answer 13

oh so
4(x^2-4x)-5(y^2+6y)=-9

Answer 14

yes, that's it... and btw, the right side should be positive 9, right?

Answer 15

oh yes! idk why I put negative.
so where do we go from here?

Answer 16

ok... this is the completing the square part...

Answer 17

help please

Answer 18

what should the value of c be in order for the polynomial to be a perfect square trinomial?

\(\large x^2 -4x + c \)

Answer 19

4?

Answer 20

HINT: take half of the x coefficient and square it. that's what c should be.

Answer 21

yes... it's 4.. so...

Answer 22

so is( x-4)^2

Answer 23

\(\large x^2-4x +\color {red}4=(x-2)^2 \)
agreed?

Answer 24

yes

Answer 25

ok... so back to the original problem...

Answer 26

4(x-2)^2-5(y+3)^2=9

Answer 27

then?

Answer 28

\(\large 4(x^2-4x+\color{red}4 )-5(y^2+6y)=9\color{red} {+ 16}\)
since you added 4 to the x group to make it a perfect square, do you see why you needed to add 16 to the right side?

Answer 29

yes

Answer 30

ok... so when you do the same for the y group, what do you need to add to the right side also?

Answer 31

-45

Answer 32

yes... so what should the right side be then?

Answer 33

-20

Answer 34

yes... lemme summarize what you have so far...

Answer 35

ok

Answer 36

\(\large 4(x^2-4x+4 )-5(y^2+6y+9)=9+ 16-45 \)
\(\large 4(x-2)^2-5(y+3)^2=-20 \)

now, standard form of a conic has a "1" on the right hand side...
divide both sides by -20. what u got?

Answer 37

(x-2)^2/5-(y+3)^2/4=1

Answer 38

\(\large 4(x-2)^2-5(y+3)^2=-20 \)
\(\large \frac{4(x-2)^2}{-20}-\frac{5(y+3)^2}{-20}=\frac{-20}{-20} \)

simplify that last line...

Answer 39

huh?

Answer 40

um.. you y group should be positive, and the x group should be negative...

Answer 41

ok..i just switched them

Answer 42

can you help we with another one similar?

Answer 43

\(\large \frac{\cancel{4}(x-2)^2}{\cancel{-20}-5}-\frac{\cancel5(y+3)^2}{\cancel{-20}-4}=1 \)

Answer 44

cleaned up it should be:
\(\large \frac{(y+3)^2}{4}-\frac{(x-2)^2}{5}=1 \)
which is an equation of a hyperbola.

Answer 45

ok thanks

Answer 46

yw... if you have a similar one, post it as a separate question and someone should help u out... :)