Question Number 198241 by sonukgindia last updated on 15/Oct/23
Answered by Frix last updated on 15/Oct/23
$$\frac{{x}^{\mathrm{2}} }{\mathrm{9}}−\frac{{y}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{1}\:\mathrm{Hyperbola},\:−\mathrm{3}\geqslant{x}\vee{x}\geqslant\mathrm{3} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{16}\:\mathrm{Circle},\:−\mathrm{3}\leqslant{x}\leqslant\mathrm{5} \\ $$$$\Rightarrow\:\mathrm{3}\:\mathrm{intersections} \\ $$$${I}_{\mathrm{1}} =\left(\frac{\mathrm{57}}{\mathrm{13}},\:\frac{\mathrm{16}\sqrt{\mathrm{3}}}{\mathrm{13}}\right)\:\:\:\:\:{I}_{\mathrm{2}} =\left(−\mathrm{3},\:\mathrm{0}\right)\:\:\:\:\:{I}_{\mathrm{3}} =\left(\frac{\mathrm{57}}{\mathrm{13}},\:−\frac{\mathrm{16}\sqrt{\mathrm{3}}}{\mathrm{13}}\right) \\ $$