Question

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

Short answer

Equation of tangent is $y=2x-3$

Step-by-step solution

Step 1. Given information  

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

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 (2,1) So we get

$1-2m=b$

Now we put b=1-2m y=mx +b

We get

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

Step 3. Calculations 

Now put $y=mx+(1-2m)$in $x^2-y^2=3$

We get

$$\begin{gathered} x^2-{(mx+1-2m)}^2=3Nowweget \\ \left(1-m^2\right)x^2+\left(4m^2-2m\right)x-4m^2+4m-4=0 \end{gathered}$$

Step 3. Solving the quadratic equation 

Now we have to solve the quadratic equation . we get

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

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

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

Step 5. Equation of tangent

Now put m=2 in equation of tangent $y=mx+(1-2m)$

We get

$y=2x-3$