## Algebra and Trigonometry 10th Edition

$3x^2+2y^2-18x -16y+58=0$ or, $3(x^2-6x+9)+2(y^2-8y+16)=-58+27+32$ or, $3(x-3)^2+2(y-4)^2=1$ or, $\dfrac{(x-3)^2}{1/3}+\dfrac{(y-4)^2}{1/2}=1$ So, the given statement is true.