Question Statement

Find the equations of the tangents and normals to the ellipse $15x^2+6y^2-8x-5y+2=0$ at the point with ordinate $\frac{1}{3}$.

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Background and Explanation

To find the equations of tangents and normals to a curve, we first identify the specific points on the curve using the given coordinate (the ordinate $y$). We then use implicit differentiation to find the derivative $\frac{dy}{dx}$, which represents the slope of the tangent line. The slope of the normal line is the negative reciprocal of the tangent's slope. Finally, we apply the point-slope form of a linear equation.

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Solution

Finding the Points of Contact

The given equation of the ellipse is: $15x^2+6y^2-8x-5y+2=0$

We are given that the ordinate is $\frac{1}{3}$, which means $y=\frac{1}{3}$. To find the corresponding $x$-coordinates, we substitute $y=\frac{1}{3}$ into equation (i):

$\begin{array}{rl}15x^2+6{\left(\frac{1}{3}\right)}^2-8x-5\left(\frac{1}{3}\right)+2& =0\\ 15x^2+6\left(\frac{1}{9}\right)-8x-\frac{5}{3}+2& =0\\ 15x^2+\frac{2}{3}-8x-\frac{5}{3}+2& =0\\ 15x^2-8x+1& =0\end{array}$

Now, we solve this quadratic equation using factorization: $\begin{array}{rl}15x^2-5x-3x+1& =0\\ 5x(3x-1)-1(3x-1)& =0\\ (5x-1)(3x-1)& =0\end{array}$

This gives us two values for $x$: $5x-1=0\Rightarrow x=\frac{1}{5}$ $3x-1=0\Rightarrow x=\frac{1}{3}$

Thus, there are two points on the ellipse where the ordinate is $\frac{1}{3}$: Point A: $\left(\frac{1}{5},\frac{1}{3}\right)$ and Point B: $\left(\frac{1}{3},\frac{1}{3}\right)$.

Finding the Slope Function

To find the slope of the tangent at any point, we differentiate equation (i) implicitly with respect to $x$:

$\begin{array}{rl}\frac{d}{dx}(15x^2)+\frac{d}{dx}(6y^2)-\frac{d}{dx}(8x)-\frac{d}{dx}(5y)+\frac{d}{dx}(2)& =0\\ 30x+12y\frac{dy}{dx}-8-5\frac{dy}{dx}& =0\\ (12y-5)\frac{dy}{dx}& =-(30x-8)\\ \frac{dy}{dx}& =-\frac{30x-8}{12y-5}\end{array}$

Equations at Point $A\left(\frac{1}{5},\frac{1}{3}\right)$

1. Slope of the tangent ($m_t$): $m_t=\frac{dy}{dx}{|}_{(\frac{1}{5},\frac{1}{3})}=-\frac{30\left(\frac{1}{5}\right)-8}{12\left(\frac{1}{3}\right)-5}=-\frac{6-8}{4-5}=-\frac{-2}{-1}=-2$

2. Equation of the tangent: Using $y-y_1=m_t(x-x_1)$: $\begin{array}{rl}y-\frac{1}{3}& =-2\left(x-\frac{1}{5}\right)\\ y-\frac{1}{3}& =-2x+\frac{2}{5}\end{array}$ Multiply by the LCD (15): $15y-5=-30x+6\Rightarrow 30x+15y-11=0$

3. Equation of the normal: The slope of the normal ($m_n$) is the negative reciprocal of $m_t$: $m_n=\frac{1}{2}$. $\begin{array}{rl}y-\frac{1}{3}& =\frac{1}{2}\left(x-\frac{1}{5}\right)\\ y-\frac{1}{3}& =\frac{1}{2}x-\frac{1}{10}\end{array}$ Multiply by the LCD (30): $30y-10=15x-3\Rightarrow 15x-30y+7=0$

Equations at Point $B\left(\frac{1}{3},\frac{1}{3}\right)$

1. Slope of the tangent ($m_t$): $m_t=\frac{dy}{dx}{|}_{(\frac{1}{3},\frac{1}{3})}=-\frac{30\left(\frac{1}{3}\right)-8}{12\left(\frac{1}{3}\right)-5}=-\frac{10-8}{4-5}=-\frac{2}{-1}=2$

2. Equation of the tangent: Using $y-y_1=m_t(x-x_1)$: $\begin{array}{rl}y-\frac{1}{3}& =2\left(x-\frac{1}{3}\right)\\ y-\frac{1}{3}& =2x-\frac{2}{3}\end{array}$ Multiply by 3: $3y-1=6x-2\Rightarrow 6x-3y-1=0$

3. Equation of the normal: The slope of the normal ($m_n$) is the negative reciprocal of $m_t$: $m_n=-\frac{1}{2}$. $\begin{array}{rl}y-\frac{1}{3}& =-\frac{1}{2}\left(x-\frac{1}{3}\right)\\ y-\frac{1}{3}& =-\frac{1}{2}x+\frac{1}{6}\end{array}$ Multiply by 6: $\begin{array}{rl}6y-2& =-3x+1\\ 3x+6y-3& =0\\ x+2y-1& =0(\text{Dividing by 3})\end{array}$

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Key Formulas or Methods Used

• Implicit Differentiation: Used to find $\frac{dy}{dx}$ for equations where $y$ is not isolated.
• Point-Slope Form: $y-y_1=m(x-x_1)$.
• Tangent Slope: $m_t=\frac{dy}{dx}$ at a specific point.
• Normal Slope: $m_n=-\frac{1}{m_t}$.
• Quadratic Factorization: Used to find $x$ values from the ordinate.

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Summary of Steps

1. Substitute $y=\frac{1}{3}$ into the ellipse equation to find the two points of contact: $A(1/5,1/3)$ and $B(1/3,1/3)$.
2. Differentiate the ellipse equation implicitly to find the general expression for the slope $\frac{dy}{dx}$.
3. Calculate the tangent slope $m_t$ at Point A and use the point-slope formula to find the tangent equation.
4. Calculate the normal slope $m_n$ at Point A and find the normal equation.
5. Repeat the process for Point B to find the second set of tangent and normal equations.