Question Statement

Find the equation of the tangent and normal to the hyperbola $6x^2-6y^2-14x+21y-5=0$ at the point $(2,3)$.

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Background and Explanation

To find the equations of the tangent and normal lines to a curve, we first use implicit differentiation to find the derivative $\frac{dy}{dx}$, which represents the slope of the tangent line at a given point. The normal line is perpendicular to the tangent, so its slope is the negative reciprocal of the tangent's slope ($m_n=-\frac{1}{m_t}$).

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Solution

The given equation of the hyperbola is: $6x^2-6y^2-14x+21y-5=0$

Tangent Line

To find the slope of the tangent, we differentiate the equation with respect to $x$: $6(2x)-6(2y)\frac{dy}{dx}-14+21\frac{dy}{dx}=0$

Rearranging the terms to solve for $\frac{dy}{dx}$: $(12y-21)\frac{dy}{dx}=12x-14$ $\frac{dy}{dx}=\frac{12x-14}{12y-21}$

Now, we evaluate the derivative at the point $(2,3)$ to find the slope ($m$): $\begin{array}{rl}\frac{dy}{dx}& =\frac{12(2)-14}{12(3)-21}\\ & =\frac{24-14}{36-21}\\ & =\frac{10}{15}=\frac{2}{3}\end{array}$

The slope of the tangent line at $(2,3)$ is $m=\frac{2}{3}$. Using the point-slope form $y-y_1=m(x-x_1)$: $\begin{array}{rl}y-3& =\frac{2}{3}(x-2)\\ 3(y-3)& =2(x-2)\\ 3y-9& =2x-4\\ 2x-3y+5& =0\end{array}$ This is the equation of the tangent line at $(2,3)$.

Normal Line

The slope of the normal line is the negative reciprocal of the tangent slope: $\begin{array}{rl}\text{Slope of normal line}& =\frac{-1}{\text{Slope of tangent}}\\ & =\frac{-1}{\left(\frac{2}{3}\right)}=\frac{-3}{2}\end{array}$

Using the point-slope form $y-y_1=m(x-x_1)$ for the normal line at $(2,3)$: $\begin{array}{rl}y-3& =\frac{-3}{2}(x-2)\\ 2(y-3)& =-3(x-2)\\ 2y-6& =-3x+6\\ 3x+2y+2& =0\end{array}$ This is the equation of the normal line.

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Key Formulas or Methods Used

• Implicit Differentiation: Used to find $\frac{dy}{dx}$ for equations where $y$ is not isolated.
• Slope of Tangent ($m_t$): The value of the first derivative $\frac{dy}{dx}$ at the specific point.
• Slope of Normal ($m_n$): Calculated as $m_n=-\frac{1}{m_t}$.
• Point-Slope Form: $y-y_1=m(x-x_1)$, used to construct the linear equations.

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Summary of Steps

1. Differentiate the hyperbola equation implicitly with respect to $x$.
2. Isolate $\frac{dy}{dx}$ to find the general expression for the slope.
3. Substitute the point $(2,3)$ into the derivative to find the specific slope of the tangent.
4. Apply the point-slope formula to find the tangent line equation.
5. Calculate the negative reciprocal of the tangent slope to find the normal slope.
6. Apply the point-slope formula again to find the normal line equation.