Answer: [tex]\dfrac{x^2}{9}+\dfrac{y^2}{9}=1[/tex]
Step-by-step explanation:
Given
[tex]x=3\sin 2\phi\\y=3\cos 2\phi[/tex]
square and add [tex]x\ \text{and}\ y[/tex] terms i.e.
[tex]\Rightarrow x^2+y^2=[3\sin 2\phi]^2+[3\cos 2\phi]^2\\\Rightarrow x^2+y^2=9\sin ^22\phi+9\cos ^2 2\phi\\\Rightarrow x^2+y^2=9[\sin^2 2\phi +\cos^2 2\phi]\\[/tex]
Using identity [tex]\sin ^2\theta +\cos ^2 \theta =1[/tex]
[tex]\Rightarrow x^2+y^2=9\\\\\Rightarrow \dfrac{x^2}{9}+\dfrac{y^2}{9}=1[/tex]
