Q. 92 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 92–98, use Descartes’s method from Problem 91 to find the equation of the line tangent to each graph at the given point.

92. $x^2+y^2=10;$ at $(1,3)$.

Step-by-Step Solution

Answer

The equation of tangent line is $x+3y-10=0$.

Step 1. Given information

The equation of tangent line must ne in the form of the $y=mx+b$.

Substitute $(x,y)=(1,3)$

$$\begin{gathered} 3=m\left(1\right)+b \\ 3=m+b \\ 3-m=b \\ b=3-m \end{gathered}$$

Now Substitute $b=3-m$ in $y=mx+b$.

$y=mx+(3-m)$.

Step 2. Substitute y = m x + 3 - m in given equation x 2 + y 2 = 10 .

We get

$$\begin{gathered} x^2+{[mx+(3-m\left)\right]}^2=10 \\ {x}^2+m^2{x}^2+2mx(3-m)+{(3-m)}^2=10 \\ {x}^2(1+m^2)+6mx-2m^{2}x+9-6m+m^2=10 \\ {x}^2(1+m^2)+x(6m-2m^2)+(m^2-6m-1)=10 \end{gathered}$$

Using the quadratic formula,

$x=\frac{-(6m-2m^2)\pm \sqrt{{(6m-2m^2)}^2-4(1+m^2)(m^2-6m-1)}}{2(1+m^2)}$

To obtain a unique solution, the two roots must be equal which means

$$\begin{gathered} {(6m-2m^2)}^2-4(1+m^2)(m^2-6m-1)=0 \\ (36m^2-24m^3+4m^4)-4(m^2-6m-1)-4(m^4-6m^3-m^2)=0 \\ 36m^2+24m+4=0 \\ 9m^2+6m+1=0 \\ {(3m+1)}^2=0m=-\frac{1}{3} \end{gathered}$$

Step 3. Substitute m = - 1 3 in y = m x + b .

We get

$$\begin{gathered} y=-\frac{1}{3}x+b \\ y=-\frac{1}{3}x+(3+\frac{1}{3}) \\ 3y=-x+10 \\ 3y+x-10=0 \\ x+3y-10=0 \end{gathered}$$

So the required equation of a tangent line of $x^2+y^2=10$ at $(1,3)$ is $x+3y-10=0$.$(1,3)$