Question Statement

Find the equations of the tangent and normal to the ellipse $2x^2+3y^2-5x-10y+5=0$ at the point $(3,2)$.

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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 derivative $\frac{dy}{dx}$ gives the slope of the tangent line ($m$), while the slope of the normal line is the negative reciprocal ($-\frac{1}{m}$) because the normal is perpendicular to the tangent.

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Solution

The equation of the ellipse is: $2x^2+3y^2-5x-10y+5=0$

Step 1: Find the derivative $\frac{dy}{dx}$

To find the slope, we differentiate the equation with respect to $x$:

$2(2x)+3(2y)\frac{dy}{dx}-5-10\frac{dy}{dx}=0$

Simplify the terms: $4x+6y\frac{dy}{dx}-5-10\frac{dy}{dx}=0$

Group the terms containing $\frac{dy}{dx}$ on one side: $(4x-5)+(6y-10)\frac{dy}{dx}=0$ $(6y-10)\frac{dy}{dx}=-(4x-5)$

Solve for $\frac{dy}{dx}$: $\frac{dy}{dx}=-\frac{4x-5}{6y-10}$

Step 2: Calculate the slope of the tangent at $(3,2)$

Substitute $x=3$ and $y=2$ into the derivative:

$\begin{array}{rl}\frac{dy}{dx}& =-\frac{4(3)-5}{6(2)-10}\\ & =-\frac{12-5}{12-10}\\ & =-\frac{7}{2}\end{array}$

The slope of the tangent line ($m$) at the point $(3,2)$ is $-\frac{7}{2}$.

Step 3: Find the equation of the tangent line

Using the point-slope form $y-y_1=m(x-x_1)$ with the point $(3,2)$ and $m=-\frac{7}{2}$:

$\begin{array}{rl}y-2& =\frac{-7}{2}(x-3)\\ 2(y-2)& =-7(x-3)\\ 2y-4& =-7x+21\\ 7x+2y-25& =0\end{array}$

The equation of the tangent line is $7x+2y-25=0$.

Step 4: Find the equation of the normal line

The slope of the normal line is the negative reciprocal of the tangent's slope. Since the tangent slope is $-\frac{7}{2}$, the normal slope is $\frac{2}{7}$.

Using the point-slope form with the point $(3,2)$ and $m=\frac{2}{7}$:

$\begin{array}{rl}y-2& =\frac{2}{7}(x-3)\\ 7(y-2)& =2(x-3)\\ 7y-14& =2x-6\\ 2x-7y+8& =0\end{array}$

The equation of the normal line is $2x-7y+8=0$.

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Key Formulas or Methods Used

• Implicit Differentiation: Used to find $\frac{dy}{dx}$ when $y$ is not explicitly isolated.
• Slope of Tangent ($m$): Evaluated as $\frac{dy}{dx}$ at the specific point $(x_1,y_1)$.
• Slope of Normal: $m_{normal}=-\frac{1}{m_{tangent}}$.
• Point-Slope Form: $y-y_1=m(x-x_1)$.

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Summary of Steps

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