Mathematics, 22.12.2019 10:31 Arealbot

Write and equation of the translated or rotated graph in general form (picture below)

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Answer from: michael1295

The answer is hyperbola; (x')² - (y')² - 16 = 0 ⇒ answer (a)

Step-by-step explanation:

* At first lets talk about the general form of the conic equation

- Ax² + Bxy + Cy² + Dx + Ey + F = 0

∵ B² - 4AC < 0 , if a conic exists, it will be either a circle or an ellipse.

∵ B² - 4AC = 0 , if a conic exists, it will be a parabola.

∵ B² - 4AC > 0 , if a conic exists, it will be a hyperbola.

* Now we will study our equation:

xy = -8

∵ A = 0 , B = 1 , C = 0

∴ B² - 4 AC = (1)² - 4(0)(0) = 1 > 0

∴ B² - 4AC > 0

∴ The graph is hyperbola

* The equation xy = -8

∵ We have term xy that means we rotated the graph about

the origin by angle Ф

∵ Ф = π/4

∴ We rotated the x-axis and the y-axis by angle π/4

* That means the point (x' , y') it was point (x , y)

- Where x' = xcosФ - ysinФ and y' = xsinФ + ycosФ

∴ x' = xcos(π/4) - ysin(π/4) , y' = xsin(π/4) + ycos(π/4)

∴ x' = x/√2 - y/√2 = (x - y)/√2

∴ y' = x/√2 + y/√2 = (x + y)/√2

* Lets substitute x' and y' in the 1st answer

∵ (x')² - (y')² - 16 = 0

∴ $(\frac{x-y}{\sqrt{2}})^{2}-(\frac{x+y}{\sqrt{2}})^{2}=$

( $\frac{x^{2}-2xy+y^{2}}{2})-(\frac{x^{2}+2xy+y^{2}}{2})-16=0$

* Lets open the bracket

∴ $\frac{x^{2}-2xy+y^{2}-x^{2}-2xy-y^{2}}{2}-16=0$

* Lets add the like terms

∴ $\frac{-4xy}{2}-16=0$

* Simplify the fraction

∴ -2xy - 16 = 0

* Divide the equation by -2

∴ xy + 8 = 0

∴ xy = -8 ⇒ our equation

∴ Answer (a) is our answer

∴ The answer is hyperbola; (x')² - (y')² - 16 = 0

* Look at the graph:

- The black is the equation (x')² - (y')² - 16 = 0

- The purple is the equation xy = -8

- The red line is x'

- The blue line is y'

Write and equation of the translated or rotated graph in general form (picture below)

Answer from: Quest

answer: 15

step-by-step explanation:

Answer from: Quest

the system of inequalities is

$y> (1/3)x$

$y\leq-x+4$

step-by-step explanation:

step 1

find the equation of the dashed line

we have the points

(0,0) and (3,1)

the line passes through the origin, so, is a direct variation

find the slope

$m=y/x=1/3$

the equation of the dashed line is

$y=(1/3)x$

the solution of the inequality a s the shaded area above the dashed line

therefore

the equation of the inequality a is

$y> (1/3)x$

step 2

find the equation of the solid line

we have the points

(4,0) and (0,4)

find the slope

$m=(4-0)/(0-4)=-1$

the equation of the solid line is

$y=-x+4$

the solution of the inequality b s the shaded area below the solid line

therefore

the equation of the inequality b is

$y\leq-x+4$

step 3

the system of inequalities is

$y> (1/3)x$

$y\leq-x+4$

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