Respuesta :
The graph of the path of the swooping bird is shown in figure below. The general form of the hyperbola equation is given by:
[tex] \frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1[/tex]
So we can order the equation of the problem by multiplying it by the following term:
[tex]\frac{1}{49000}[/tex]
Therefore:
[tex]\frac{y^{2}}{35^{2}}-\frac{x^{2}}{2^{2}}=1[/tex]
∴ [tex]\frac{y^{2}}{1225}-\frac{x^{2}}{4}=1[/tex]
Given that the origin lies at ground level, the bird is closest to the ground at the vertices of the parable, that is, when x = 0 (this will give us two solutions, but we will take the positive value because the bird flight over the air)
[tex]\frac{y^{2}}{1225}-\frac{0^{2}}{4}=1 \rightarrow y=\sqrt{1225} \rightarrow \boxed{height=35m}[/tex]
[tex] \frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1[/tex]
So we can order the equation of the problem by multiplying it by the following term:
[tex]\frac{1}{49000}[/tex]
Therefore:
[tex]\frac{y^{2}}{35^{2}}-\frac{x^{2}}{2^{2}}=1[/tex]
∴ [tex]\frac{y^{2}}{1225}-\frac{x^{2}}{4}=1[/tex]
Given that the origin lies at ground level, the bird is closest to the ground at the vertices of the parable, that is, when x = 0 (this will give us two solutions, but we will take the positive value because the bird flight over the air)
[tex]\frac{y^{2}}{1225}-\frac{0^{2}}{4}=1 \rightarrow y=\sqrt{1225} \rightarrow \boxed{height=35m}[/tex]
