Question

Use Descartes’s method to find the equation of the line tangent to the graph $2x^2+3y^2=14$at the given point (1,2).

Short answer

Equation of tangent is $y=-\frac{1}{3}x+\frac{7}{3}$

Step-by-step solution

Step 1. Given information  

$2x^2+3y^2=14$ we have to find tangent at (1,2)

Step 2. General equation of tangent Now we put b=2-m  y=mx +bWe get

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

$2-m=b$

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 $2x^2+3y^2=14$

We get

$$\begin{gathered} 2x^2+3{(mx+2-m)}^2=142x^2+3m^2{x}^2+6mx-3m^{2}x+6mx+12-6m-3m^{2}x-6m+3m^2=14 \\ \left(2+3m^2\right)x^2+\left(12m-6m^2\right)x+3m^2-12m-2-0 \end{gathered}$$

Step 3. Solving the quadratic equation

Now we have to solve the quadratic equation . we get

$x=\frac{\left(6m^2-12m\right)\pm \sqrt{{\left(6m^2-12m\right)}^2-4\left(2+3m^2\right)\left(3m^2-12m-2\right)}}{2\left(2+3m^2\right)}$

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

$$\begin{gathered} {\left.(6m^2-12m\right)}^2-4\left(2+3m^2\right)\left(3m^2-12m-2\right)=0 \\ (3m+1{)}^2=0 \\ m=\frac{-1}{3} \end{gathered}$$

Step 5. Equation of tangent

Now put $m=\frac{-1}{3}$in equation of tangent $y=mx+(2-m)$

We get

$y=-\frac{1}{3}x+\frac{7}{3}$