Question Statement

Find the equation of the tangent to the circle $36x^2+36y^2-72y+11=0$ at the point on the circle with abscissa $\frac{1}{2}$.

---

Background and Explanation

To find the equation of a tangent line, we first need the coordinates of the point(s) of tangency and the slope of the curve at those points. The "abscissa" refers to the $x$-coordinate. We use implicit differentiation to find the derivative $\frac{dy}{dx}$, which represents the slope of the tangent line.

---

Solution

Part (a): Finding the points of tangency

The given equation of the circle is: $36x^2+36y^2-72y+11=0$

The abscissa of the point is given as $x=\frac{1}{2}$. To find the corresponding $y$-coordinates, we substitute $x=\frac{1}{2}$ into equation (i):

$\begin{array}{rl}36{\left(\frac{1}{2}\right)}^2+36y^2-72y+11& =0\\ 9+36y^2-72y+11& =0\\ 36y^2-72y+20& =0\end{array}$

Dividing the entire equation by 4 to simplify: $9y^2-18y+5=0$

We solve this quadratic equation for $y$ using the quadratic formula $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$:

$\begin{array}{rl}y& =\frac{-(-18)\pm \sqrt{(-18{)}^2-4(9)(5)}}{2(9)}\\ y& =\frac{18\pm \sqrt{324-180}}{18}\\ y& =\frac{18\pm \sqrt{144}}{18}\\ y& =\frac{18\pm 12}{18}\end{array}$

This gives us two possible values for $y$:

1. $y=\frac{18-12}{18}=\frac{6}{18}=\frac{1}{3}$
2. $y=\frac{18+12}{18}=\frac{30}{18}=\frac{5}{3}$

Thus, there are two points on the circle with an abscissa of $\frac{1}{2}$: $A\left(\frac{1}{2},\frac{1}{3}\right)$ and $B\left(\frac{1}{2},\frac{5}{3}\right)$.

Part (b): Finding the slope of the tangent

To find the slope, we differentiate equation (i) implicitly with respect to $x$:

$\begin{array}{rl}36(2x)+36\left(2y\frac{dy}{dx}\right)-72\frac{dy}{dx}+0& =0\\ 72x+72y\frac{dy}{dx}-72\frac{dy}{dx}& =0\\ 72\left(x+y\frac{dy}{dx}-\frac{dy}{dx}\right)& =0\\ x+(y-1)\frac{dy}{dx}& =0\\ \frac{dy}{dx}& =-\frac{x}{y-1}\\ \frac{dy}{dx}& =\frac{x}{1-y}\end{array}$

Part (c): Equation of the tangent at point $A\left(\frac{1}{2},\frac{1}{3}\right)$

First, calculate the slope $m$ at point $A$: $m=\frac{dy}{dx}=\frac{\frac{1}{2}}{1-\frac{1}{3}}=\frac{\frac{1}{2}}{\frac{2}{3}}=\frac{3}{4}$

Using the point-slope form $y-y_1=m(x-x_1)$: $\begin{array}{rl}y-\frac{1}{3}& =\frac{3}{4}\left(x-\frac{1}{2}\right)\\ y-\frac{1}{3}& =\frac{3}{4}x-\frac{3}{8}\\ \frac{3}{4}x-y+\frac{1}{3}-\frac{3}{8}& =0\end{array}$

Multiply the entire equation by 24 to clear the denominators: $18x-24y+8-9=0$ $18x-24y-1=0$

Part (d): Equation of the tangent at point $B\left(\frac{1}{2},\frac{5}{3}\right)$

First, calculate the slope $m$ at point $B$: $m=\frac{dy}{dx}=\frac{\frac{1}{2}}{1-\frac{5}{3}}=\frac{\frac{1}{2}}{-\frac{2}{3}}=-\frac{3}{4}$

Using the point-slope form $y-y_1=m(x-x_1)$: $\begin{array}{rl}y-\frac{5}{3}& =-\frac{3}{4}\left(x-\frac{1}{2}\right)\\ y-\frac{5}{3}& =-\frac{3}{4}x+\frac{3}{8}\\ \frac{3}{4}x+y-\frac{5}{3}-\frac{3}{8}& =0\end{array}$

Multiply the entire equation by 24 to clear the denominators: $18x+24y-40-9=0$ $18x+24y-49=0$

---

Key Formulas or Methods Used

• Abscissa: The $x$-coordinate of a point.
• Quadratic Formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
• Implicit Differentiation: Used to find $\frac{dy}{dx}$ for equations where $y$ is not explicitly isolated.
• Point-Slope Form of a Line: $y-y_1=m(x-x_1)$

---

Summary of Steps

1. Substitute $x=\frac{1}{2}$ into the circle's equation to find the corresponding $y$ values.
2. Solve the resulting quadratic equation to identify the two points of tangency: $A(1/2,1/3)$ and $B(1/2,5/3)$.
3. Differentiate the circle equation implicitly to find the general expression for the slope $\frac{dy}{dx}$.
4. Plug the coordinates of point $A$ into the derivative to find the slope, then use the point-slope formula to find the first tangent equation.
5. Repeat the process for point $B$ to find the second tangent equation.