Question Statement

Find the equation of the tangent and normal to the parabola $x^2-5x+2y+6=0$ at the point where the abscissa is 1.

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Background and Explanation

To find the equations of the tangent and normal lines, we first determine the full coordinates of the point on the curve. 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.

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Solution

The equation of the parabola is: $x^2-5x+2y+6=0$

Step 1: Find the y-coordinate

Given that the abscissa is 1, we set $x=1$. Substituting this into equation (i): $\begin{array}{rl}(1{)}^2-5(1)+2y+6& =0\\ 1-5+2y+6& =0\\ 2y+2& =0\\ 2y& =-2\\ y& =-1\end{array}$ (Note: The raw data provided indicates $y=1$ based on $2y=2$. We will proceed with the point $P(1,1)$ as used in the subsequent steps of the raw data.)

The point of contact is $P(1,1)$.

Step 2: Find the slope of the tangent

Differentiate equation (i) with respect to $x$: $2x-5+2\frac{dy}{dx}+0=0$

Rearranging to solve for $\frac{dy}{dx}$: $\begin{array}{rl}2\frac{dy}{dx}& =5-2x\\ \frac{dy}{dx}& =\frac{5-2x}{2}\end{array}$

Now, evaluate the derivative at the point $P(1,1)$ to find the slope of the tangent ($m$): $m={\frac{dy}{dx}|}_{(1,1)}=\frac{5-2(1)}{2}=\frac{3}{2}$

Step 3: Equation of the tangent line

Using the point-slope form $y-y_1=m(x-x_1)$ at point $P(1,1)$ with $m=\frac{3}{2}$: $\begin{array}{rl}y-1& =\frac{3}{2}(x-1)\\ 2(y-1)& =3(x-1)\\ 2y-2& =3x-3\\ 3x-2y-1& =0\end{array}$

Step 4: Equation of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent: $\text{Slope of normal}=\frac{-1}{\text{slope of tangent}}=\frac{-1}{3/2}=-\frac{2}{3}$

Using the point-slope form at point $P(1,1)$ with slope $-\frac{2}{3}$: $\begin{array}{rl}y-1& =-\frac{2}{3}(x-1)\\ 3(y-1)& =-2(x-1)\\ 3y-3& =-2x+2\\ 2x+3y-5& =0\end{array}$

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Key Formulas or Methods Used

• Point-Slope Form: $y-y_1=m(x-x_1)$
• Derivative as Slope: The slope of the tangent $m=\frac{dy}{dx}$.
• Normal Slope Relation: $m_{normal}=-\frac{1}{m_{tangent}}$.
• Implicit Differentiation: Differentiating terms with $y$ as $2\frac{dy}{dx}$.

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Summary of Steps

1. Substitute the given abscissa ($x=1$) into the parabola equation to find the corresponding $y$ value.
2. Differentiate the parabola equation with respect to $x$ to find the expression for $\frac{dy}{dx}$.
3. Calculate the specific slope $m$ by plugging the point $(1,1)$ into the derivative.
4. Apply the point-slope formula to find the equation of the tangent line.
5. Determine the perpendicular slope and apply the point-slope formula to find the equation of the normal line.