Chapter 3.5, Problem 52E

To determine

To find: The equation of the both tangent lines to the ellipse passing through the point.

Answer to Problem 52E

The equation of the tangent line to the curve passing through the point $\left(12,3\right)$ at $\left(0,3\right)$ and $\left(\frac{24}{5},\frac{-9}{5}\right)$ are $y=3$ and $y=\frac{2}{3}x-5$.

Explanation of Solution

Given:

The equation of the ellipse $x^2+4y^2=36$.

The point is $\left(12,3\right)$.

Derivative rules:

Chain rule: $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$.

Calculation:

Obtain the slope of tangent to the equation.

Differentiate $x^2+4y^2=36$ implicitly with respect to x.

$\begin{array}{l}\frac{d}{dx}\left(x^2+4y^2\right)=\frac{d}{dx}\left(36\right)\\ \frac{d}{dx}\left(x^2\right)+4\frac{d}{dx}\left(y^2\right)=0\\ 2x+4\frac{d}{dx}\left(y^2\right)=0\end{array}$

Apply the chain rule and simplify the terms.

$\begin{array}{l}2x+4\left[\frac{d}{dy}\left(y^2\right)\frac{dy}{dx}\right]=0\\ 2x+4\left[2y\frac{dy}{dx}\right]=0\\ 2x+8y\frac{dy}{dx}=0\\ \frac{dy}{dx}=\frac{-x}{4y}\end{array}$

Thus, the derivative of the equation of the ellipse is $\frac{dy}{dx}=\frac{-x}{4y}$.

That is, the slope of the tangent is $\frac{dy}{dx}=\frac{-x}{4y}$.

Let $\left(a,b\right)$ be a point on the curve.

The slope of tangent to the curve at $\left(a,b\right)$ is $\frac{dy}{dx}=\frac{-a}{4b}$.

The equation of the tangent line passing through the point $\left(12,3\right)$ and the slope $\frac{-a}{4b}$ is computed as follows.

$\begin{array}{l}y-3=\frac{-a}{4b}\left(x-12\right)\\ y-3=\frac{-ax}{4b}+\frac{12a}{4b}\\ 4by-12b=-ax+12a\end{array}$

$ax+4by=12a+12b$ (1)

Here, the tangent line is also passing through the point $\left(a,b\right)$.

Substitute $\left(a,b\right)$ for $\left(x,y\right)$,

$a\left(a\right)+4b\left(b\right)=12a+12b$

$a^2+4b^2=12\left(a+b\right)$ (2)

The value of the curve $x^2+4y^2=36$ at $\left(a,b\right)$ is $a^2+4b^2=36$ (3)

Substitute the equation (3) in equation (2).

$\begin{array}{l}12\left(a+b\right)=36\\ a+b=3\\ b=3-a\end{array}$

Substitute $b=3-a$ in equation (3).

$\begin{array}{l}{a}^2+4{\left(3-a\right)}^2=36\\ a^2+4\left(9+a^2-6a\right)=36\\ a^2+36+4a^2-24a-36=0\\ 5a^2-24a=0\end{array}$

Simplify the quadratic equation,

$\begin{array}{l}a\left(5a-24\right)=0\\ a=0\text{ or }a=\frac{24}{5}\end{array}$

Substitute $a=0$ in $b=3-a$,

Thus, the value of b is 3.

Substitute $a=\frac{24}{5}$ in $b=3-a$.

$\begin{array}{c}b=3-\frac{24}{5}\\ =\frac{15-24}{5}\\ =\frac{-9}{5}\end{array}$

Thus, the points are $\left(0,3\right)\text{ and }\left(\frac{24}{5},\frac{-9}{5}\right)$.

Substitute $\left(0,3\right)$ for $\left(a,b\right)$ in equation (1),

$\begin{array}{l}\left(0\right)x+4y\left(3\right)=12\left(0\right)+12\left(3\right)\\ 12y=36\\ y=3\end{array}$

Thus, the equation of the tangent line to the curve passing through the point $\left(12,3\right)$ at $\left(0,3\right)$ is $y=3$.

Substitute $\left(\frac{24}{5},\frac{-9}{5}\right)$ for $\left(a,b\right)$ in equation (1).

$\begin{array}{l}\left(\frac{24}{5}\right)x+4y\left(\frac{-9}{5}\right)=12\left(\frac{24}{5}\right)+12\left(\frac{-9}{5}\right)\\ 24x-36y=288-108\\ 24x-36y=180\\ y=\frac{2}{3}x-5\end{array}$

Thus, the equation of the tangent line to the curve passing through the point $\left(12,3\right)$ at $\left(\frac{24}{5},\frac{-9}{5}\right)$ is $y=\frac{2}{3}x-5$.