Chapter 11.6, Problem 92AYU

To determine

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

Answer to Problem 92AYU

Solution:

$y = - \frac{1}{3}x + \frac{10}{3}$

Explanation of Solution

Given:

$x^2 + y^2 = 10$   –––(Eq1)

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, $x^2 + y^2 = 10$

$\begin{array}{l}{x}^2 + m^2{x}^2 + 9 + m^2 - 6m + 6mx - 2m^{2}x = 10\\ (1 + m^2)x^2 + (6m - 2m^2)x + m^2 - 6m - 1 = 0\end{array}$

Solving the above equation,

$x = \frac{ - (6m - 2m^2) \pm \sqrt{{(6m - 2m^2)}^2 - 4(1 + m^2)(m^2 - 6m - 1)}}{2(1 + m^2)}$

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

$\begin{array}{rrr}\hfill {(6m - 2m^2)}^2 - 4(1 + m^2)(m^2 - 6m - 1)& \hfill = & 0\hfill \\ \hfill 9m^2 + 6m + 1& \hfill = & 0\hfill \\ \hfill m& \hfill = & \hfill - \frac{1}{3}\end{array}$

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

$m = - \frac{1}{3},b = 3 - m = \frac{10}{3}$

Therefore, $y = - \frac{1}{3}x + \frac{10}{3}$.