Question

The equation $3x^2+4y^2-18x+16y+43=c$.

• A. Cannot represent a real pair of straight lines for any value of $c$
• B. Represent an ellipse, if $c>0$
• C. Represent empty set, if $c<0$
• D. A point, if $c=0$

Answer 1

The correct option is A Cannot represent a real pair of straight lines for any value of $c$

$3x^2+4y^2-18x+16y+43-c=0$

Here, $\begin{array}{cc}a& =3\\ h& =0\\ b& =4\\ g& =-9\\ f& =8\\ c& =43-c\end{array}$

$\begin{array}{cc}\Delta & =\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\mid \\ & =\begin{array}{ccc}3& 0& -9\\ 0& 4& 8\\ -9& 8& 43-4\end{array}\\ & =3\left(4\right(43-c)-64)-9(0+36)\\ & =12(43-6)-192-324\\ & =516-12c-516\\ & =-12c\end{array}$

If $c>0$, then $△\ne 0$

and $h^2<$ ab $($ also true $)$

$0<12$

it must represent an ellipse, if $C>0$