Question

verify that the given point is on the curve and find the lines that are a. tangent and b. normal to the curve at the given point.
4x² - √7xy + y² = 36, (√7,8)
the point is on the curve because when √7 is substituted for x and 8 is substituted for y, the resulting statement is 364 = 36, which is a false statement. (simplify your answers.)

Explanation:

Step1: Substitute x and y values

Substitute \(x = \sqrt{7}\) and \(y = 8\) into \(4x^{2}-\sqrt{7}xy + y^{2}\).
\[

$$\begin{align*} &4(\sqrt{7})^{2}-\sqrt{7}\times\sqrt{7}\times8+8^{2}\\ =&4\times7 - 7\times8+64\\ =&28-56 + 64\\ =&36 \end{align*}$$

\]

Step2: Implicit - differentiate the equation

Differentiate \(4x^{2}-\sqrt{7}xy + y^{2}=36\) with respect to \(x\).
Using the sum - difference rule and product rule:
The derivative of \(4x^{2}\) is \(8x\).
For \(-\sqrt{7}xy\), by product rule \((uv)^\prime=u^\prime v+uv^\prime\) where \(u = -\sqrt{7}x\) and \(v = y\), we get \(-\sqrt{7}y-\sqrt{7}x\frac{dy}{dx}\).
The derivative of \(y^{2}\) using the chain - rule is \(2y\frac{dy}{dx}\).
So, \(8x-\sqrt{7}y-\sqrt{7}x\frac{dy}{dx}+2y\frac{dy}{dx}=0\).
Solve for \(\frac{dy}{dx}\):
\[

$$\begin{align*} -\sqrt{7}x\frac{dy}{dx}+2y\frac{dy}{dx}&=\sqrt{7}y - 8x\\ \frac{dy}{dx}(2y-\sqrt{7}x)&=\sqrt{7}y - 8x\\ \frac{dy}{dx}&=\frac{\sqrt{7}y - 8x}{2y-\sqrt{7}x} \end{align*}$$

\]
Substitute \(x = \sqrt{7}\) and \(y = 8\) into \(\frac{dy}{dx}\):
\[

$$\begin{align*} \frac{dy}{dx}&=\frac{\sqrt{7}\times8 - 8\times\sqrt{7}}{2\times8-\sqrt{7}\times\sqrt{7}}\\ &=\frac{8\sqrt{7}-8\sqrt{7}}{16 - 7}\\ &=0 \end{align*}$$

\]

Step3: Find the tangent line

The equation of a line in point - slope form is \(y - y_{1}=m(x - x_{1})\), where \((x_{1},y_{1})=(\sqrt{7},8)\) and \(m = 0\) (the slope of the tangent line).
So, \(y-8=0(x - \sqrt{7})\), and the tangent line is \(y = 8\).

Step4: Find the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line \(m_{t}=0\), the slope of the normal line is undefined.
The equation of the normal line is \(x=\sqrt{7}\).

Answer:

a. The equation of the tangent line is \(y = 8\).
b. The equation of the normal line is \(x=\sqrt{7}\).