Question

(a) Use implicit differentiation to find an equation of the tangent line to the hyperbola \(\frac{x^{2}}{6}-\frac{y^{2}}{8}=1\) at (3,-2) (b) Show that the equation of the tangent line to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) at \(\left(x_{0}, y_{0}\right)\) is \(\frac{x_{0} x}{a^{2}}-\frac{y_{0} y}{b^{2}}=1\)

Short answer

The equation of the tangent line to the hyperbola \(\frac{x^{2}}{6}-\frac{y^{2}}{8}=1\) at \((3,-2)\) is \(2x + y = 8\). Furthermore, the equation of the tangent line to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) at \((x_{0}, y_{0})\) is indeed \(\frac{x_{0} x}{a^{2}}-\frac{y_{0} y}{b^{2}}=1\).

Step-by-step solution

Differentiate the given equation

First, differentiate the equation of the hyperbola. For part (a), implicit differentiation of the given equation \(\frac{x^{2}}{6}-\frac{y^{2}}{8}=1\) results in \(\frac{2x}{6} - \frac{2y y'}{8} = 0\). For part (b), differentiate \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) to get \(\frac{2x x'}{a^{2}} - \frac{2y y'}{b^{2}}= 0\). Here, \(y'\) represents the derivative of \(y\) with respect to \(x\), also known as the slope of the tangent line.

Find the slope

In part (a), rearrange the above equation to the form \(y' = y'(x)\) by isolating \(y'\) on one side: \(y' = \frac{x}{3y}\). Substitute \(x=3\) and \(y=-2\) to get \(y' = -\frac{1}{2}\), which is the slope of the tangent. In part (b), similarly rearrange the equation from step 1 to get \(y' = \frac{x x'}{a^{2}}\). This expression represents the slope of the tangent line at any point \((x, y)\) on the hyperbola.

Find the equation of the tangent line

In part (a), use the point-slope form of a line: \(y - y_{1} = m(x - x_{1})\), where \(m\) is the slope and \((x_{1}, y_{1})\) is the given point on the curve. Substitute \(m= -\frac{1}{2}\), \(x_{1}=3\) and \(y_{1}=-2\) to get \(y - (-2) = -\frac{1}{2} (x - 3)\), which simplifies to \(2x + y = 8\). This is the equation of the required tangent line. For part (b), derive \(\frac{x_{0} x}{a^{2}}-\frac{y_{0} y}{b^{2}}=1\) from the slope-expression in step 2 by substituting \(x'=1\), \(x=x_{0}\) and \(y=y_{0}\).