Chapter 11.6, Problem 93AYU

To determine

To find: Equation of tangent line using Descartes’ method.

Answer to Problem 93AYU

Solution:

$y = 2x + 1$

Explanation of Solution

Given:

$y = x^2 + 2$   –––(Eq 1)

And the point $(1,3)$

Calculation:

Equation of tangent line is $y = mx + b$

Using the fact that the point $(1,3)$ is on the line, we get

$\begin{array}{lll}b\hfill & = \hfill & 3 - m\hfill \\ y\hfill & = \hfill & mx + (3 - m)\hfill \end{array}$

Eq 1 implies, $y = x^2 + 2$

$\begin{array}{l}mx + (3 - m) = x^2 + 2\hfill \\ x^2 - mx - 1 + m = 0\hfill \end{array}$

Solving the above equation,

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

Using Descartes’ method to obtain unique solution, discriminant $= 0$

$\begin{array}{rrr}\hfill m^2 - 4(m - 1)& \hfill = & 0\hfill \\ \hfill m^2 - 4m + 4& \hfill = & 0\hfill \\ \hfill m& \hfill = & 2\hfill \end{array}$

Equation of tangent line is $y = mx + b$

$m = 2,b = 3 - 2 = 1$

Therefore, $y = 2x + 1$.