Chapter 11: Q. 92 (page 775)

In Problems 92–98, use Descartes’s method from Problem 91 to find the equation of the line tangent to each graph at the given point.
92. $x^{2} + y^{2} = 10;$ at $(1, 3)$ .

Short Answer

Expert verified

The equation of tangent line is $x + 3 y - 10 = 0$ .

Step by step solution

Step 1. Given information

The equation of tangent line must ne in the form of the $y = m x + b$ .

Substitute $(x, y) = (1, 3)$

$3 = m (1) + b 3 = m + b 3 - m = b b = 3 - m$

Now Substitute $b = 3 - m$ in $y = m x + b$ .

$y = m x + (3 - m)$ .

Step 2. Substitute y=mx+3-m in given equation x2+y2=10.

We get

$x^{2} + {[m x + (3 - m)]}^{2} = 10 x^{2} + m^{2} x^{2} + 2 m x (3 - m) + {(3 - m)}^{2} = 10 x^{2} (1 + m^{2}) + 6 m x - 2 m^{2} x + 9 - 6 m + m^{2} = 10 x^{2} (1 + m^{2}) + x (6 m - 2 m^{2}) + (m^{2} - 6 m - 1) = 10$

Using the quadratic formula,

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

To obtain a unique solution, the two roots must be equal which means

${(6 m - 2 m^{2})}^{2} - 4 (1 + m^{2}) (m^{2} - 6 m - 1) = 0 (36 m^{2} - 24 m^{3} + 4 m^{4}) - 4 (m^{2} - 6 m - 1) - 4 (m^{4} - 6 m^{3} - m^{2}) = 0 36 m^{2} + 24 m + 4 = 0 9 m^{2} + 6 m + 1 = 0 {(3 m + 1)}^{2} = 0 m = - \frac{1}{3}$

Step 3. Substitute m=-13 in y=mx+b.

We get

$y = - \frac{1}{3} x + b y = - \frac{1}{3} x + (3 + \frac{1}{3}) 3 y = - x + 10 3 y + x - 10 = 0 x + 3 y - 10 = 0$

So the required equation of a tangent line of $x^{2} + y^{2} = 10$ at $(1, 3)$ is $x + 3 y - 10 = 0$ . $(1, 3)$

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In Problems 92–98, use Descartes’s method from Problem 91 to find the equation of the line tangent to each graph at the given point.
92. $x^{2} + y^{2} = 10;$ at $(1, 3)$ .

Short Answer

Step by step solution

Step 1. Given information

Step 2. Substitute y=mx+3-m in given equation x2+y2=10.

Step 3. Substitute m=-13 in y=mx+b.

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In Problems 92–98, use Descartes’s method from Problem 91 to find the equation of the line tangent to each graph at the given point.92.x2+y2=10;at(1,3).

Short Answer

Step by step solution

Step 1. Given information

Step 2. Substitute y=mx+3-m in given equation x2+y2=10.

Step 3. Substitute m=-13 in y=mx+b.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Calculus

Statistics

Mechanics Maths

Theoretical and Mathematical Physics

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

In Problems 92–98, use Descartes’s method from Problem 91 to find the equation of the line tangent to each graph at the given point.
92. $x^{2} + y^{2} = 10;$ at $(1, 3)$ .