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 $$y = \square$$.

Explanation:

Step1: Differentiate the equation implicitly

Differentiate \(\frac{x^{2}}{9}+\frac{y^{2}}{25} = 1\) with respect to \(x\).
Using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), we have \(\frac{2x}{9}+\frac{2y y^\prime}{25}=0\).
Then solve for \(y^\prime\):

$$
LATEXBLOCK0
$$

Step2: Find the slope of the tangent at the given point

Substitute \(x = 1\) and \(y=\frac{10}{3}\sqrt{2}\) into \(y^\prime\).
\(y^\prime=-\frac{25\times1}{9\times\frac{10}{3}\sqrt{2}}=-\frac{25}{30\sqrt{2}}=-\frac{5}{6\sqrt{2}}=-\frac{5\sqrt{2}}{12}\)

Step3: Use the point - slope form \(y - y_{0}=m(x - x_{0})\)

Here \(x_{0}=1,y_{0}=\frac{10}{3}\sqrt{2},m =-\frac{5\sqrt{2}}{12}\)

$$
LATEXBLOCK1
$$

Answer:

\(y =-\frac{5\sqrt{2}}{12}x+\frac{15\sqrt{2}}{4}\)