Question Statement

Find the equation of the tangent and normal to the parabola $2y^2-3y+11x-16=0$ at the point $(1,-1)$.

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Background and Explanation

To find the equations of the tangent and normal lines, we first need to determine the slope of the curve at the given point using implicit differentiation. The slope of the tangent line is equal to the derivative $\frac{dy}{dx}$, while the slope of the normal line is the negative reciprocal of the tangent's slope.

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Solution

Step 1: Differentiate the equation

To find the slope, we differentiate the equation $2y^2-3y+11x-16=0$ with respect to $x$:

$4y\frac{dy}{dx}-3\frac{dy}{dx}+11=0$

Next, we group the $\frac{dy}{dx}$ terms to solve for the derivative:

$(4y-3)\frac{dy}{dx}=-11$

$\frac{dy}{dx}=\frac{-11}{4y-3}$

Step 2: Find the slope of the tangent at $(1,-1)$

We substitute the $y$-coordinate of the given point $(1,-1)$ into our derivative formula:

$\begin{array}{rl}\frac{dy}{dx}& =\frac{-11}{4(-1)-3}\\ & =\frac{-11}{-7}=\frac{11}{7}\end{array}$

Thus, the slope of the tangent line ($m$) at $(1,-1)$ is $\frac{11}{7}$.

Step 3: Find the equation of the tangent line

Using the point-slope form $y-y_1=m(x-x_1)$ with the point $(1,-1)$ and slope $m=\frac{11}{7}$:

$\begin{array}{rl}y-(-1)& =\frac{11}{7}(x-1)\\ 7(y+1)& =11(x-1)\\ 7y+7& =11x-11\\ 11x-7y-18& =0\end{array}$

The equation of the tangent line is $11x-7y-18=0$.

Step 4: Find the slope of the normal line

The normal line is perpendicular to the tangent line. Therefore, its slope is the negative reciprocal of the tangent's slope:

$\begin{array}{rl}\text{Slope of normal line}& =\frac{-1}{\text{Slope of tangent}}\\ & =\frac{-1}{\left(\frac{11}{7}\right)}=-\frac{7}{11}\end{array}$

Step 5: Find the equation of the normal line

Using the point-slope form again with the same point $(1,-1)$ and the new slope $m=-\frac{7}{11}$:

$\begin{array}{rl}y-(-1)& =\frac{-7}{11}(x-1)\\ 11(y+1)& =-7(x-1)\\ 11y+11& =-7x+7\\ 7x+11y+4& =0\end{array}$

The equation of the normal line is $7x+11y+4=0$.

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Key Formulas or Methods Used

• Implicit Differentiation: Used to find $\frac{dy}{dx}$ when $y$ is not explicitly defined as a function of $x$.
• Slope of Tangent: $m=\frac{dy}{dx}$ evaluated at the specific point.
• Point-Slope Form: $y-y_1=m(x-x_1)$.
• Perpendicular Slopes: The slope of the normal $m_n=-\frac{1}{m_t}$.

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Summary of Steps

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