Question

57: Show that the ellipse \({x^2}/{a^2} + {y^2}/{b^2} = 1\)and the hyperbola \({x^2}/{A^2} - {y^2}/{{\bf{B}}^2} = 1\)are orthogonal trajectories if \({A^2} < {a^2}\) and \({a^2} - {b^2} = {A^2} + {B^2}\) (so the ellipse and hyperbola have the same foci).

Short answer

It is proved that ellipse and hyperbola are orthogonal trajectories if \({A^2} < {a^2}\), and\({a^2} - {b^2} = {A^2} + {B^2}\).

Step-by-step solution

Graph the curves

Plot the curves of ellipse and hyperbola. The graph is shown below.

Use Implicit differentiation to find the slope of both curves

Differentiate both sides of the equation with respect to \(x\).

\(\begin{aligned}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} &= 1\\\frac{{2x}}{{{a^2}}} + \frac{{2yy'}}{{{b^2}}} &= 0\\\frac{{2yy'}}{{{b^2}}} &= - \frac{{2x}}{{{a^2}}}\\\frac{{yy'}}{{{b^2}}} &= - \frac{x}{{{a^2}}}\\y' &= - \frac{{x{b^2}}}{{y{a^2}}}\\{m_1} &= - \frac{{x{b^2}}}{{y{a^2}}}\;\;\;\; \ldots \left( 1 \right)\end{aligned}\)

And,

\(\begin{aligned}{c}\frac{{{x^2}}}{{{A^2}}} - \frac{{{y^2}}}{{{B^2}}} &= 1\\\frac{{2x}}{{{A^2}}} - \frac{{2yy'}}{{{B^2}}} &= 0\\\frac{{2yy'}}{{{B^2}}} &= \frac{{2x}}{{{A^2}}}\\\frac{{yy'}}{{{B^2}}} &= \frac{x}{{{A^2}}}\\y' &= \frac{{x{B^2}}}{{y{A^2}}}\\{m_2} &= \frac{{x{B^2}}}{{y{A^2}}}\;\;\;\; \ldots \left( 2 \right)\end{aligned}\)

Find the value of \({m_1}{m_2}\).

Substitute the above slope values into \({m_1}{m_2}\) and simplify.

\(\begin{aligned}{m_1}{m_2} &= - \frac{{x{b^2}}}{{y{a^2}}} \cdot \frac{{x{B^2}}}{{y{A^2}}}\\ &= - \frac{{{b^2}{B^2}}}{{{a^2}{A^2}}} \cdot \frac{{{x^2}}}{{{y^2}}}\;\; \ldots \left( 3 \right)\end{aligned}\)

Simplify further the expression

Subtract the equation \(\frac{{{x^2}}}{{{A^2}}} - \frac{{{y^2}}}{{{B^2}}} = 1\) from \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) to determine the value of \(\frac{{{x^2}}}{{{y^2}}}\).

\(\begin{aligned}{c}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \left( {\frac{{{x^2}}}{{{A^2}}} - \frac{{{y^2}}}{{{B^2}}}} \right) &= 1 - 1\\\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{x^2}}}{{{A^2}}} + \frac{{{y^2}}}{{{B^2}}} &= 0\\\frac{{{y^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{B^2}}} &= \frac{{{x^2}}}{{{A^2}}} - \frac{{{x^2}}}{{{a^2}}}\\\frac{{{y^2}{B^2} + {y^2}{b^2}}}{{{b^2}{B^2}}} &= \frac{{{x^2}{a^2} - {x^2}{A^2}}}{{{A^2}{a^2}}}\\\frac{{{y^2}\left( {{B^2} + {b^2}} \right)}}{{{b^2}{B^2}}} &= \frac{{{x^2}\left( {{a^2} - {A^2}} \right)}}{{{A^2}{a^2}}}\end{aligned}\)

Use the given condition \({a^2} - {b^2} = {A^2} + {B^2}\), that is, \({a^2} - {A^2} = {b^2} + {B^2}\)to simplify.

\(\begin{aligned}{c}\frac{{{y^2}\left( {{B^2} + {b^2}} \right)}}{{{b^2}{B^2}}} &= \frac{{{x^2}\left( {{B^2} + {b^2}} \right)}}{{{A^2}{a^2}}}\\\frac{{{y^2}}}{{{b^2}{B^2}}} &= \frac{{{x^2}}}{{{A^2}{a^2}}}\\\frac{{{x^2}}}{{{y^2}}} &= \frac{{{A^2}{a^2}}}{{{b^2}{B^2}}}\end{aligned}\)

Substitute the values

Substitute \(\frac{{{x^2}}}{{{y^2}}} = \frac{{{A^2}{a^2}}}{{{b^2}{B^2}}}\) into equation (3).

\(\begin{aligned}{c}{m_1}{m_2} &= - \frac{{{b^2}{B^2}}}{{{a^2}{A^2}}} \cdot \left( {\frac{{{A^2}{a^2}}}{{{b^2}{B^2}}}} \right)\\ &= - 1\end{aligned}\)

As \({m_1}{m_2} = - 1\), this implies that hyperbola and ellipse are orthogonal.