As a result of these equations to the canonical form. According to the schedule, the first - it is an ellipse. But reduced to canonical form? $$(x^2+4y^2-4)\sqrt y = 0 $$ $$3y+2\sqrt{9-x^2} = 0 $$
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in the first one, why have you written $\ast \sqrt y$ – Will Jagy Jan 29 '17 at 21:36
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it is so important ?? – Frip Jan 29 '17 at 21:42
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well, if the product of two things is required to be zero, then one or the other is zero, maybe both. Also, the usual interpretation is the demand for the square root to be real. So, it is not so much an ellipse as it is half an ellipse (with $y>0$) together with the entire $x$ axis. – Will Jagy Jan 29 '17 at 21:46