Question

e. find the equation of the normal to the curve ( y^{2}-x^{2}=5 ) at point ( (2,3) ).
hint: use implicit differentiation

Explanation:

Step1: Differentiate the equation implicitly

Differentiate \(y^{2}-x^{2}=5\) with respect to \(x\).
Using the chain - rule \(\frac{d}{dx}(y^{2}) = 2y\frac{dy}{dx}\) and \(\frac{d}{dx}(x^{2})=2x\).
So, \(2y\frac{dy}{dx}-2x = 0\).
Solve for \(\frac{dy}{dx}\):

$$
LATEXBLOCK0
$$

Step2: Find the slope of the tangent at the point \((2,3)\)

Substitute \(x = 2\) and \(y = 3\) into \(\frac{dy}{dx}\).
\(\frac{dy}{dx}\big|_{(2,3)}=\frac{2}{3}\)

Step3: Find the slope of the normal

The slope of the normal \(m_{n}\) is the negative reciprocal of the slope of the tangent.
If the slope of the tangent \(m_{t}=\frac{2}{3}\), then \(m_{n}=-\frac{3}{2}\)

Step4: Use the point - slope form of a line

The point - slope form of a line is \(y - y_{1}=m(x - x_{1})\), where \((x_{1},y_{1})=(2,3)\) and \(m =-\frac{3}{2}\)

$$
LATEXBLOCK1
$$

Answer:

\(y=-\frac{3}{2}x + 6\)