1. The problem gives three equations: $$y^2=4ax$$, $$x^2=4ay$$, and $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$. 2. These represent conic sections: the first two are parabolas, and the third is an ellipse. 3. To analyze or solve these, we use the standard forms of conic sections and their properties. 4. For the parabola $$y^2=4ax$$, the vertex is at the origin, and it opens rightward if $$a>0$$. 5. For the parabola $$x^2=4ay$$, the vertex is at the origin, and it opens upward if $$a>0$$. 6. The ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ has center at the origin, with semi-major axis $$a$$ and semi-minor axis $$b$$. 7. If the problem is to find intersections or relations among these curves, substitute and solve accordingly. 8. For example, to find intersection points between the two parabolas, set $$y^2=4ax$$ and $$x^2=4ay$$. 9. From $$y^2=4ax$$, express $$x=\frac{y^2}{4a}$$. 10. Substitute into $$x^2=4ay$$: $$\left(\frac{y^2}{4a}\right)^2=4ay$$ $$\frac{y^4}{16a^2}=4ay$$ Multiply both sides by $$16a^2$$: $$y^4=64a^3 y$$ 11. Rearranged: $$y^4 - 64a^3 y=0$$ $$y(y^3 - 64a^3)=0$$ 12. So, $$y=0$$ or $$y^3=64a^3$$. 13. Thus, $$y=0$$ or $$y=4a$$. 14. For $$y=0$$, from $$y^2=4ax$$, $$0=4ax$$ implies $$x=0$$. 15. For $$y=4a$$, from $$y^2=4ax$$: $$ (4a)^2=4a x$$ $$16 a^2=4a x$$ $$x=4a$$. 16. So intersection points are $$ (0,0) $$ and $$ (4a,4a) $$. 17. The ellipse equation can be analyzed separately or used to find intersections similarly. Final answer: The two parabolas intersect at points $$ (0,0) $$ and $$ (4a,4a) $$.