Question

What is the nature of the curve $3x^2-4y^2+7=0$

Answer 1

The equation of the given curve is

$F(x,y)=3x^2-4y^2+7=0$

Here $F(x,-y)=3x^2-4{(-y)}^2+7=3x^2-4y^2+7=0=F(x,y)$

$F(-x,y)=3{(-x)}^2-4y^2+7=3x^2-4y^2+7=0=F(x,y)$

$F(-x,-y)=3{(-x)}^2-4{(-y)}^2+7=3x^2-4y^2+7=0=F(x,y)$

$\therefore$ the given curve is symmetrical about the $x-$axis,$y-$axis and about the origin.