Question

find the lines that are (a) tangential and (b) normal to each curve at the given point.

\\( \frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 25 } = 1, \left( 1, \frac { 10 } { 3 } \sqrt { 2 } \
ight) \\)

the equation of the tangent line to the point \\( \left( 1, \frac { 10 } { 3 } \sqrt { 2 } \
ight) \\) is \\( y = - \frac { 5 \sqrt { 2 } } { 12 } x + \frac { 15 \sqrt { 2 } } { 4 } \\).

the equation of the normal line to the point \\( \left( 1, \frac { 10 } { 3 } \sqrt { 2 } \
ight) \\) is \\( y = \square \\).

Explanation:

Step1: Find the slope of the normal line

The slope of the tangent line \(m_{tangent}=-\frac{5\sqrt{2}}{12}\).
Since the normal line is perpendicular to the tangent line, if two lines with slopes \(m_1\) and \(m_2\) are perpendicular, then \(m_1\times m_2=- 1\). Let the slope of the normal line be \(m_{normal}\), so \(m_{normal}=\frac{12}{5\sqrt{2}}=\frac{6\sqrt{2}}{5}\).

Step2: Use the point - slope form \(y - y_0=m(x - x_0)\)

We have the point \((x_0,y_0)=(1,\frac{10\sqrt{2}}{3})\) and \(m = \frac{6\sqrt{2}}{5}\).
Substitute into the point - slope form: \(y-\frac{10\sqrt{2}}{3}=\frac{6\sqrt{2}}{5}(x - 1)\).
Expand the right - hand side: \(y-\frac{10\sqrt{2}}{3}=\frac{6\sqrt{2}}{5}x-\frac{6\sqrt{2}}{5}\).
Then \(y=\frac{6\sqrt{2}}{5}x-\frac{6\sqrt{2}}{5}+\frac{10\sqrt{2}}{3}\).
Find a common denominator (15): \(y=\frac{6\sqrt{2}\times3}{15}x-\frac{6\sqrt{2}\times3}{15}+\frac{10\sqrt{2}\times5}{15}\).
\(y=\frac{18\sqrt{2}}{15}x-\frac{18\sqrt{2}}{15}+\frac{50\sqrt{2}}{15}\).
Simplify: \(y=\frac{6\sqrt{2}}{5}x+\frac{32\sqrt{2}}{15}\).

Answer:

\(y=\frac{6\sqrt{2}}{5}x+\frac{32\sqrt{2}}{15}\)