# What is the equation of the normal to the curve x^2 + y^2 = 25 at (4, 3)?

**Problem**

What is the equation of the normal to the curve $x^2 + y^2 = 25$ at (4, 3)?

A. $4x + 3y = 0$ | C. $3x + 4y = 0$ |

B. $3x - 4y = 0$ | D. $4x - 3y = 0$ |

**Answer Key**

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**Solution**

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**Solution by Analytic Geometry**

$xx_1 + yy_1 = 25$ ← equation of tangent at any point on the curve

At (4, 3), the equation of tangent is

$4x + 3y = 25$

Note:

$Ax + By + C_1 = 0$ and $Bx - Ay + C_2 = 0$ are perpendicular lines.

Hence, the equation of normal through (4, 3) is

$3x - 4y = 3(4) - 4(3)$

$3x - 4y = 0$ ← *answer*

**Solution by Calculus with Analytic Geometry**

$2x + 2y \, y' = 0$

$y' = -\dfrac{x}{y}$ ← slope of the tangent at any point

The slope of normal is therefore

$m = \dfrac{y}{x}$

At point (4, 3)

$m = \dfrac{3}{4}$

Equation of normal

$y - y_1 = m(x - x_1)$

$y - 3 = \frac{3}{4}(x - 4)$

$4y - 12 = 3x - 12$

$3x - 4y = 0$ ← *answer*

**Solution by Drawing**

From the figure, the equation of normal through (4, 3) is...

$y = mx + b$

$y = \dfrac{3}{4}x + 0$

$4y = 3x$

$3x - 4y = 0$ ← *answer*

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