Chapter 2.4, Problem 36E

To determine

To find: The equations of all lines tangent to $y=9-x^2$ that pass through the point $\left(1,12\right)$.

Answer to Problem 36E

The equations of all lines tangent to $y=9-x^2$ that pass through the point $\left(1,12\right)$ are $y=2x+10$ and $y=-6x+18$.

Explanation of Solution

Given information:

The curve is $y=9-x^2$ and the passing point is $\left(1,12\right)$.

Calculation:

The formula for the slope of curve at $x=a$ is equal to $\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$.

Substitute $a+h$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=9-x^2\\ f\left(a+h\right)=9-{\left(a+h\right)}^2\\ f\left(a+h\right)=9-\left(a^2+h^2+2ah\right)\\ f\left(a+h\right)=9-a^2-h^2-2ah\end{array}$

Substitute $a$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=9-x^2\\ f\left(a\right)=9-a^2\end{array}$

Substitute $9-a^2-h^2-2ah$ for $f\left(a+h\right)$ and $9-a^2$ for $f\left(a\right)$ in the formula for slope.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\underset{h\to 0}{\lim }\frac{9-a^2-h^2-2ah-\left(9-a^2\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{-h^2-2ah}{h}\\ =\underset{h\to 0}{\lim }\frac{h\left(-h-2a\right)}{h}\\ =\underset{h\to 0}{\lim }\left(-h-2a\right)\end{array}$

Further simplify.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=0-2a\\ =-2a\end{array}$

The tangent line to the curve $y=9-x^2$ at the point $x=a$.

  $\begin{array}{c}y-f\left(a\right)=m\left(x-a\right)\\ y-\left(9-a^2\right)=-2a\left(x-a\right)\\ y-9+a^2=-2ax+2a^2\\ y=-2ax+a^2+9\end{array}$

As the tangent line passes through the point $\left(1,12\right)$ . Substitute $1$ for $x$ and $12$ for $y$ in the tangent line.

  $\begin{array}{c}y=-2ax+a^2+9\\ 12=-2a\left(1\right)+a^2+9\\ 12=-2a+a^2+9\\ a^2-2a-3=0\end{array}$

Further simplify,

  $\begin{array}{c}{a}^2-2a-3=0\\ a^2-3a+a-3=0\\ a\left(a-3\right)+1\left(a-3\right)=0\\ \left(a+1\right)\left(a-3\right)=0\end{array}$

So, the values of $a$ are $-1$ and $3$.

Substitute $-1$ for $a$ in the tangent line.

  $\begin{array}{c}y=-2ax+a^2+9\\ y=-2\left(-1\right)x+{\left(-1\right)}^2+9\\ y=2x+1+9\\ y=2x+10\end{array}$

Substitute $3$ for $a$ in the tangent line.

  $\begin{array}{c}y=-2ax+a^2+9\\ y=-2\left(3\right)x+{\left(3\right)}^2+9\\ y=-6x+9+9\\ y=-6x+18\end{array}$

Therefore, the equations of all lines tangent to $y=9-x^2$ that pass through the point $\left(1,12\right)$ are $y=2x+10$ and $y=-6x+18$.