Given:
[tex]\frac{(x^{\prime})^2}{15}-\frac{(y^{\prime})^2}{6}=1\text{ at }\theta=30^o[/tex]To find:
We need to find an equation for the conic in the xy-plane.
Explanation:
We can find the conic equation by using the following equation.
[tex]x^{\prime}=x\cos \theta+y\sin \theta\text{ and }y^{\prime}=y\cos \theta-x\sin \theta[/tex][tex]\text{Substitute }\theta=30^o\text{ in the eqatuions.}[/tex][tex]x^{\prime}=x\cos 30^o+y\sin 30^o\text{ and }y^{\prime}=y\cos 30^o-x\sin 30^o\text{.}[/tex][tex]\text{Use }\cos 30^o=\frac{\sqrt[]{3}}{2}\text{ and }\sin 30^o=\frac{1}{2}\text{.}[/tex][tex]x^{\prime}=x(\frac{\sqrt[]{3}}{2})+y(\frac{1}{2})\text{ and }y^{\prime}=y(\frac{\sqrt[]{3}}{2})-x(\frac{1}{2})[/tex][tex]x^{\prime}=\frac{\sqrt[]{3}}{2}x+\frac{1}{2}y\text{ and }y^{\prime}=\frac{\sqrt[]{3}}{2}y-\frac{1}{2}x[/tex][tex]\text{ Substitute }x^{\prime}=\frac{\sqrt[]{3}}{2}x+\frac{1}{2}y\text{ and }y^{\prime}=\frac{\sqrt[]{3}}{2}y-\frac{1}{2}x\text{ in the given equation.}[/tex][tex]\frac{(\frac{\sqrt[]{3}}{2}x+\frac{1}{2}y)^2}{15}-\frac{(\frac{\sqrt[]{3}}{2}y-\frac{1}{2}x)^2}{6}=1[/tex][tex]\frac{1}{15}(\frac{\sqrt[]{3}}{2}x+\frac{1}{2}y)^2-\frac{1}{6}(\frac{\sqrt[]{3}}{2}y-\frac{1}{2}x)^2=1[/tex][tex]\frac{1}{15}\mleft\lbrace(\frac{\sqrt[]{3}x}{2})^2+(2\times\frac{\sqrt[]{3}x}{2}\times\frac{y}{2})+(\frac{y}{2})^2\mright\rbrace-\frac{1}{6}\mleft\lbrace(\frac{\sqrt[]{3}y}{2})^2-2\times\frac{\sqrt[]{3}y}{2}\times\frac{x}{2}+(\frac{x}{2})^2\mright\rbrace=1[/tex][tex]\frac{1}{15}\mleft\lbrace\frac{3x}{4}^2+\frac{\sqrt[]{3}xy}{2}+\frac{y^2}{4}^{}\mright\rbrace-\frac{1}{6}\mleft\lbrace\frac{3y}{4}^2-\frac{\sqrt[]{3}xy}{2}+\frac{x}{4}^2\mright\rbrace=1[/tex][tex]\frac{3x^2}{15\times4}^{}+\frac{\sqrt[]{3}xy}{15\times2}+\frac{y^2}{15\times4}^{}-\frac{3y^2}{6\times4}^{}+\frac{\sqrt[]{3}xy}{6\times2}-\frac{x}{6\times4}^2=1[/tex][tex]\frac{x^2}{20}^{}+\frac{\sqrt[]{3}xy}{30}+\frac{y^2}{60}^{}-\frac{y^2}{8}^{}+\frac{\sqrt[]{3}xy}{12}-\frac{x^2}{24}^{}=1[/tex]Here LCM is 360, making the denominator 360.
[tex]18x^2+12\sqrt[]{3}xy+6y^2-45y^2+30\sqrt[]{3}xy-15x^2=360[/tex][tex]3x^2+42\sqrt[]{3}xy-39y^2-360=0[/tex]Final answer:
The equation for the conic in the xy-plane is
[tex]3x^2+42\sqrt[]{3}xy-39y^2-360=0[/tex]