Question

Write the equation in terms of a rotated x'y'-system using the angle of rotation. Write the equation involving x' and y' in standard form.

19x2 + 241/3 xy - 5y2 - 527 = 0; 0 = 30°

Answer 1

Answer:The answer is The answer is ""

Explanation:

Answer 2

The answer is "
`\bold{(x^(')^2)/(17)-(y^(')^2)/(31)=1}`"

Explanation:

In the given question there is some mistyping error if the given value is this. so, its solution can be defined as follows:

`19x^2 + 24√(3) xy - 5y^2 - 527 = 0...................(1) \\\\ \theta = 30^(\circ)`

Formula:

`\bold{x=x'\cos \theta -y' \sin \theta}`
`_(where) \ \ \ \ \theta =30`

`=x' \cos 30- y' \sin 30\\\\ =x' (√(3))/(2)- y' (1)/(2)\\\\ =(1)/(2)(x' √(3)- y')....(a)\\\\`

`\bold{y=x'\sin \theta +y' \cos \theta}`
`_(where) \ \ \ \ \theta =30`

`=x' \sin 30+y' \cos 30\\\\ =x' (1)/(2)+y' (√(3))/(2)\\\\ =(1)/(2)(x'+ y'√(3))....(b)\\\\`

put the equation (a) and equation (b) value in equation 1:

equation:

`\to 19x^2 + 24√(3) xy - 5y^2 - 527 = 0...................(1) \\\\`

`\to 19((1)/(2)(x'√(3)- y'))^2+24√(3)((1)/(2)(x'√(3)- y))( (1)/(2)(x'+y'√(3)))- 5((1)/(2)(x'+y'√(3)))^2-527= 0\\`

`\to (19)/(4)(3x^(')^2+ y^(')^(2)-2√(3)x' y')+6√(3)(√(3)x^(')^(2)+2x'y'-√(3)y^(')^2)-(5)/(4)(x^(')^2+ 3y^(')^(2)-2√(3)x' y')-527=0`

`\to (57)/(4)x^(')^2+(19)/(4)y^(')^2-(19 √(3))/(2)x^(')^2y^(')^2+18x^(')^2+12√(3)x^(')^2y^(')^2-18y^(')^2-(5)/(4)x^(')^2-(15)/(4)y^(')^2-(5√(3))/(2)x^(')^2y^(')^2-527=0\\\\\to 31x^(')^2-17y^(')^2-527=0\\\\\to 31x^(')^2-17y^(')^2=527\\\\\to (x^(')^2)/(17)-(y^(')^2)/(31)=1\\\\`