## Precalculus (6th Edition) Blitzer

The orbital body and comet can intersect at $\left( 0,-4 \right),\left( -2,0 \right),\left( 2,0 \right)$
Let us consider the paths followed by the orbital body and the comet \begin{align} & 16{{x}^{2}}+4{{y}^{2}}=64\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\ & y={{x}^{2}}-4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \end{align} Put the value of $y={{x}^{2}}-4$ into equation (I): \begin{align} & 16{{x}^{2}}+4{{\left( {{x}^{2}}-4 \right)}^{2}}=64 \\ & 16{{x}^{2}}+4\left( {{x}^{4}}-8{{x}^{2}}+16 \right)=64 \\ & 16{{x}^{2}}+4{{x}^{4}}-32{{x}^{2}}+64=64 \\ & 4{{x}^{4}}-16{{x}^{2}}=0 \end{align} Further solving it, \begin{align} & {{x}^{4}}-4{{x}^{2}}=0 \\ & {{x}^{2}}\left( {{x}^{2}}-4 \right)=0 \\ & {{x}^{2}}=0 \\ & x=0 \end{align} Substitute the value of $x=0,2,-2$ in equation (I): If, \begin{align} & x=0 \\ & y={{\left( 0 \right)}^{2}}-4 \\ & =-4 \end{align} If, \begin{align} & x=2 \\ & y={{\left( 2 \right)}^{2}}-4 \\ & =0 \end{align} If, \begin{align} & x=-2 \\ & y={{\left( -2 \right)}^{2}}-4 \\ & =0 \end{align} Thus, the possible points of intersection are $\left( 0,-4 \right),\left( -2,0 \right),\left( 2,0 \right)$.