If ax^{2}+2hxy+by^{2}+2gx+2fy+c=0 represents | vedclass

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(C) The given hyperbola is $16x^{2}-9y^{2}=144$. Dividing by $144$,we get $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$. Here,$a^{2}=9$ and $b^{2}=16$. The ecc

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$81$

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(C) The given hyperbola is $16x^{2}-9y^{2}=144$. Dividing by $144$,we get $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$. Here,$a^{2}=9$ and $b^{2}=16$. The eccentricity $e$ is given by $b^{2}=a^{2}(e^{2}-1)$,so $16=9(e^{2}-1)$,which gives $e^{2}-1=\frac{16}{9}$,so $e^{2}=\frac{25}{9}$,$e=\frac{5}{3}$. The directrices of a hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ are $x=\pm \frac{a}{e}$. Here,$a=3$ and $e=\frac{5}{3}$,so $x=\pm \frac{3}{5/3} = \pm \frac{9}{5}$. Thus,$5x-9=0$ and $5x+9=0$. The joint equation is $(5x-9)(5x+9)=0$,which is $25x^{2}-81=0$. Comparing this with $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$,we have $a=25, h=0, b=0, g=0, f=0, c=-81$. Therefore,$g+f-c = 0+0-(-81) = 81$.

If $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ represents a joint equation of the directrices of the hyperbola $16x^{2}-9y^{2}=144$,then $g+f-c=$

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