Respuesta :
Answer:
The answer is "[tex]\bold{\frac{x^{'}^2}{17}-\frac{y^{'}^2}{31}=1}[/tex]"
Step-by-step explanation:
In the given question there is some mistyping error if the given value is this. so, its solution can be defined as follows:
[tex]19x^2 + 24\sqrt{3} xy - 5y^2 - 527 = 0...................(1) \\\\ \theta = 30^{\circ}[/tex]
Formula:
[tex]\bold{x=x'\cos \theta -y' \sin \theta}[/tex] [tex]_{where} \ \ \ \ \theta =30[/tex]
[tex]=x' \cos 30- y' \sin 30\\\\ =x' \frac{\sqrt{3}}{2}- y' \frac{1}{2}\\\\ =\frac{1}{2}(x' \sqrt{3}- y')....(a)\\\\[/tex]
[tex]\bold{y=x'\sin \theta +y' \cos \theta}[/tex] [tex]_{where} \ \ \ \ \theta =30[/tex]
[tex]=x' \sin 30+y' \cos 30\\\\ =x' \frac{1}{2}+y' \frac{\sqrt{3}}{2}\\\\ =\frac{1}{2}(x'+ y'\sqrt{3})....(b)\\\\[/tex]
put the equation (a) and equation (b) value in equation 1:
equation:
[tex]\to 19x^2 + 24\sqrt{3} xy - 5y^2 - 527 = 0...................(1) \\\\[/tex]
[tex]\to 19(\frac{1}{2}(x'\sqrt{3}- y'))^2+24\sqrt{3}(\frac{1}{2}(x'\sqrt{3}- y))( \frac{1}{2}(x'+y'\sqrt{3}))- 5(\frac{1}{2}(x'+y'\sqrt{3}))^2-527= 0\\[/tex]
[tex]\to \frac{19}{4}(3x^{'}^2+ y^{'}^{2}-2\sqrt{3}x' y')+6\sqrt{3}(\sqrt{3}x^{'}^{2}+2x'y'-\sqrt{3}y^{'}^2)-\frac{5}{4}(x^{'}^2+ 3y^{'}^{2}-2\sqrt{3}x' y')-527=0[/tex]
[tex]\to \frac{57}{4}x^{'}^2+\frac{19}{4}y^{'}^2-\frac{19 \sqrt{3}}{2}x^{'}^2y^{'}^2+18x^{'}^2+12\sqrt{3}x^{'}^2y^{'}^2-18y^{'}^2-\frac{5}{4}x^{'}^2-\frac{15}{4}y^{'}^2-\frac{5\sqrt{3}}{2}x^{'}^2y^{'}^2-527=0\\\\\to 31x^{'}^2-17y^{'}^2-527=0\\\\\to 31x^{'}^2-17y^{'}^2=527\\\\\to \frac{x^{'}^2}{17}-\frac{y^{'}^2}{31}=1\\\\[/tex]
