Question

x2 − 2xy + y2 + 2x + 2y = 0. factor

Answer 1

x'2 = 8y' – 2

y'2 = 4x

x2 = 4y − 2

y'2 = 2x

Answer 2

What would the equation of the parabola be in terms of the rotated axes?

Answer 3

@e.mccormick

Answer 4

@dan815

Answer 5

@Michele_Laino

Answer 6

@sleepyjess

Answer 7

@KendrickLamar2014

Answer 8

I chopped the question up so people would come

Answer 9

@EclipsedStar

Answer 10

n engineer designing a roller coaster is using an advanced software application to test the strength of the structure. A part of the structure has a parabolic shape with the equation x2 − 2xy + y2 + 2x + 2y = 0. However, the software doesn't recognize any equation involving an xy term. What would the equation of the parabola be in terms of the rotated axes?

Answer 11

please wait I try

Answer 12

yea take your time

Answer 13

here, we have the subsequent matrix equation:
\[\left( {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{1/\sqrt 2 }&{1/\sqrt 2 } \\
{ - 1/\sqrt 2 }&{1/\sqrt 2 }
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{x'} \\
{y'}
\end{array}} \right)\]
where x' and y' are the new coordinates.
Namely we have to compute the eigenvalues and the eigenvectors of the quadratic part of your parabola

Answer 14

so we can write:
\[\begin{gathered}
x = \frac{{x' + y'}}{{\sqrt 2 }} \hfill \\
y = \frac{{x' - y'}}{{\sqrt 2 }} \hfill \\
\end{gathered} \]
those are the requested equation of our rotation

Answer 15

Now we have to substitute those equations into your parabola,
in order to rewrite your parabola in term of x' and y',so what do you get?

Answer 16

substitute

\[x=\frac{ x'+y }{ \sqrt{2} } and y=\frac{ x' -y'}{ \sqrt{2} }\]

x2 − 2xy + y2 + 2x + 2y = 0

Answer 17

that?

Answer 18

yes!

Answer 19

how do you do that?

Answer 20

x = 1/4, y = -3/4

Answer 21

you wont me to substitute the factor in to the equation

Answer 22

the solution?

Answer 23

for example:
\[\large \begin{gathered}
{x^2} + {y^2} - 2xy + 2x + 2y = \hfill \\
\hfill \\
= {\left( {\frac{{x' + y'}}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{{x' - y'}}{{\sqrt 2 }}} \right)^2} - 2\left( {\frac{{x' + y'}}{{\sqrt 2 }}} \right) \cdot \left( {\frac{{x' - y'}}{{\sqrt 2 }}} \right) + \hfill \\
+ 2\frac{{x' + y'}}{{\sqrt 2 }} + 2\frac{{x' - y'}}{{\sqrt 2 }} = ...? \hfill \\
\end{gathered} \]

Answer 24

I got this:
\[\Large y{'^2} + \sqrt 2 x' = 0\]
please check my computations

Answer 25

y'^2 = 2x?

Answer 26

I think:
y'2=-sqrt(2)x'

Answer 27

so its D

Answer 28

please can you check your options?

Answer 29

It is D:)

Answer 30

ok!