Question

3.6. Applications Tangent Line \& Normal Line Equations Example 1: Find the equation of tangent and normal to the circle \( x^{2}+y^{2}=5 \), at the point \( (2,3) \). Solution 1: \( y=m x+b \) where \( m= \) slope, \( b=y \)-intercept \[ \begin{array}{l} y=\left(5-x^{2}\right)^{\frac{1}{2}} \\ y^{\prime}=\frac{1}{2}\left(5-x^{2}\right)^{-\frac{1}{2}}(-2 x) \\ y^{\prime}(x=2)=\frac{1}{2}\left(5-(2)^{2}\right)^{-\frac{1}{2}}(-2(2))=-2 \\ y=-2 x+b \Rightarrow 3=-2(2)+b \Rightarrow \therefore b=7 \\ \therefore y=-2 x+7 \end{array} \] 2005

          3.6. Applications

Tangent Line \& Normal Line Equations
Example 1:
Find the equation of tangent and normal to the circle \( x^{2}+y^{2}=5 \), at the point \( (2,3) \).
Solution 1: \( y=m x+b \) where \( m= \) slope, \( b=y \)-intercept
\[
\begin{array}{l}
y=\left(5-x^{2}\right)^{\frac{1}{2}} \\
y^{\prime}=\frac{1}{2}\left(5-x^{2}\right)^{-\frac{1}{2}}(-2 x) \\
y^{\prime}(x=2)=\frac{1}{2}\left(5-(2)^{2}\right)^{-\frac{1}{2}}(-2(2))=-2 \\
y=-2 x+b \Rightarrow 3=-2(2)+b \Rightarrow \therefore b=7 \\
\therefore y=-2 x+7
\end{array}
\]
2005

3.6. Applications

Tangent Line & Normal Line Equations
Example 1:
Find the equation of tangent and normal to the circle x^2+y^2=5, at the point (2,3).
Solution 1: y=m x+b where m= slope, b=y-intercept

y=(5-x^2)^(1)/(2)

y^'=(1)/(2)(5-x^2)^-(1)/(2)(-2 x)

y^'(x=2)=(1)/(2)(5-(2)^2)^-(1)/(2)(-2(2))=-2

y=-2 x+b ⇒ 3=-2(2)+b ⇒∴ b=7
∴ y=-2 x+7

2005

Added by Gregg M.

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