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

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

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

y=mx +b

As this line is passes through (2,3) So we get

$$\begin{gathered} 3=m\left(2\right)+b \\ b=3-2m \end{gathered}$$

Now we put b=3-2m y=mx +b

We get

$y=mx+(3-2m)$

Now put $y=mx+(3-2m)in2y^2-x^2=14$

We get

localid="1646989878289" $$\begin{gathered} 2(mx+(3-2m){)}^2-x^2)=14 \\ 2m^2{x}^2+4(mx\left)\right(3-2m)+2(3-2m{)}^2-x^2=14 \\ {x}^2\left(2m^2-1\right)+12mx-8m^{2}x+18-24m+8m^2=14 \\ {x}^2\left(2m^2-1\right)+x\left(12m-8m^2\right)+\left(8m^2-24m+4\right)=0 \end{gathered}$$

Now we have to solve the quadratic equation . we get

$x=\frac{-\left(12m-8m^2\right)\pm \sqrt{{\left(12m-8m^2\right)}^2-4\left(2m^2-1\right)\left(8m^2-24m+4\right)}}{2\left(2m^2-1\right)}$

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

$$\begin{gathered} {\left(12m-8m^2\right)}^2-4\left(2m^2-1\right)\left(8m^2-24m+4\right)=0 \\ \left(144m^2-192m^3+64m^4\right)-4\left(16m^4-48m^3+8m^2\right)+48m^2-24m+4=0 \\ 144m^2-192m^3+64m^4-64m^4+192m^3-32m^2+32m^2-96m+16=0 \\ 144m^2-96m+16=0 \end{gathered}$$

Now we get

$$\begin{gathered} (3m-1{)}^2=0 \\ m=\frac{1}{3} \end{gathered}$$

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

We get

$3y=x+7$