Q. 96 - Precalculus Enhanced with Graphing Utilities

Question

Use Descartes’s method  to find the equation of the line tangent to the graph $3x^2+y^2=7$ at the given point (-1,2). 

Step-by-Step Solution

Answer

Equation of tangent is $2 y=3 x+7$

Step 1. Given information

$3x^2+y^2=7$  we have to find tangent at (-1,2)

Step 2. General equation of tangent

As we know that the equation of the tangent line is in the form of 

y=mx +b 

As this line is passes through (-1,2) So we get

$b=2+m$

 Now we put b=2+m  y=mx +b

We get

$y=m x+(2+m)$

Step 3. Calculations

Now put $y=m x+(2+m)$ in $3x^2+y^2=7$

We get

$$\begin{gathered} 3x^2+[mx+(2+m){]}^2=7 \\ 3x^2+m^2{x}^2+2(mx\left)\right(2+m)+(2+m{)}^2=7 \\ {x}^2\left(3+m^2\right)+x\left(4m+2m^2\right)+\left(m^2+4m-3\right)=0 \end{gathered}$$

Step 3. Solving the quadratic equatiom

Now we have to solve the quadratic equation . we get 

$x=\frac{-\left(4m+2m^2\right)\pm \sqrt{{\left(4m+2m^2\right)}^2-4\left(3+m^2\right)\left(m^2+4m-3\right)}}{2\left(3+m^2\right)}$

Now we want the unique solutions for this both the roots should be equal .So 

$$\begin{gathered} {\left(4m+2m^2\right)}^2-4\left(3+m^2\right)\left(m^2+4m-3\right)=0 \\ 16m^2+16m^3+4m^4-12m^2-48m+36-4m^4-16m^3+12m^2=0 \\ 16m^2-48m+36=0 \\ (2m-3{)}^2=0 \\ m=\frac{3}{2} \end{gathered}$$

Step 5. Equation of tangent

Now put $m=\frac{3}{2}$ in equation of tangent $y=m x+(2+m)$

We get 

$2 y=3 x+7$