Question

Show by implicit differentiation that the tangent line to the hyperbola

\(\frac{{{x^{\bf{2}}}}}{{{a^{\bf{2}}}}} - \frac{{{y^{\bf{2}}}}}{{{b^{\bf{2}}}}} = {\bf{1}}\)

at the point \(\left( {{x_{\bf{0}}},{y_{\bf{0}}}} \right)\).

Short answer

The tangent to the hyperbola is given by the equation \(\frac{{x{x_0}}}{{{a^2}}} - \frac{{y{y_0}}}{{{b^2}}} = 1\).

Step-by-step solution

Differentiate the equation \(\frac{{{x^{\bf{2}}}}}{{{a^{\bf{2}}}}} - \frac{{{y^{\bf{2}}}}}{{{b^{\bf{2}}}}} = {\bf{1}}\)

Differentiate the equation \(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\)with respect to \(x\)on both sides.

\(\begin{aligned}\frac{{2x}}{{{a^2}}} - \frac{{2yy'}}{{{b^2}}} &= 0\\y' &= \frac{{{b^2}x}}{{{a^2}y}}\end{aligned}\)

Theslope of the tangentto the ellipse at the point \(\left( {{x_0},{y_0}} \right)\) is:

\(y'\left( {{x_0},{y_0}} \right) = \frac{{{b^2}{x_0}}}{{{a^2}{y_0}}}\)

Find the equation of the tangent at \(\left( {{x_{\bf{0}}},{y_{\bf{0}}}} \right)\)

The equation of the tangent at the point \(\left( {{x_0},{y_0}} \right)\) is:

\(\begin{aligned}y - {y_0} &= \frac{{{b^2}{x_0}}}{{{a^2}{y_0}}}\left( {x - {x_0}} \right)\\\frac{{y{y_0} - y_0^2}}{{{b^2}}} &= \frac{{x{x_0} - x_0^2}}{{{a^2}}}\\\frac{{x{x_0}}}{{{a^2}}} - \frac{{y{y_0}}}{{{b^2}}} &= \frac{{x_0^2}}{{{a^2}}} - \frac{{y_0^2}}{{{b^2}}}\\\frac{{x{x_0}}}{{{a^2}}} - \frac{{y{y_0}}}{{{b^2}}} &= 1\end{aligned}\)

Thus, the tangent to the hyperbola is given by the equation \(\frac{{x{x_0}}}{{{a^2}}} - \frac{{y{y_0}}}{{{b^2}}} = 1\).