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.

94.$x^2+y=5$; at$(-2,1)$.

Short answer

The equation of the tangent line is$4x-y+9=0$.

Step-by-step solution

Step 1. Given information

The general form of the equation of the tangent line must be $y=mx+b$.

Substitute $(-2,1)$in $y=mx+b$.

$$\begin{gathered} 1=m(-2)+b \\ 1+2m=b \\ b=2m+1 \end{gathered}$$

Substitute $b=2m+1$in $y=mx+b$.

$y=mx+(2m+1)$

So the system of equations here is

$$\begin{gathered} y=mx+(2m+1) \\ {x}^2+y=5 \end{gathered}$$.

Step 2.  Solve the system of equations.

Substitute $y=mx+(2m+1)$in $x^2+y=5$.

role="math" localid="1646976038805" $$\begin{gathered} x^2+mx+(2m+1)=5 \\ {x}^2+mx+(2m-4)=0 \end{gathered}$$

Using the formula for quadratic equation,

role="math" $$\begin{gathered} x=\frac{-m\pm \sqrt{m^2-4\left(1\right)(2m-4)}}{2\left(1\right)} \\ =\frac{-m\pm \sqrt{m^2-4(2m-4)}}{2} \end{gathered}$$

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

$$\begin{gathered} m^2-4(2m-4)=0 \\ {m}^2-8m+16=0 \\ {(m-4)}^2=0m-4=0 \\ m=4 \end{gathered}$$

Step 3. Substitute m=4 in y=mx+b.

We get

$$\begin{gathered} y=4x+(2\times 4+1) \\ y=4x+9 \\ 4x-y+9=0 \end{gathered}$$

So the equation of the tangent line for$x^2+y=5$at$(-2,1)$is$4x-y+9=0$.