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).

Short answer

Equation of tangent is$2y=3x+7$

Step-by-step solution

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

$y=mx+(2+m)$

Step 3. Calculations

Now put $y=mx+(2+m)$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$y=mx+(2+m)$

We get

$2y=3x+7$