verify that the given point is on the curve and find the lines that are a. tangent and b. normal to the…

Explanation:

Step1: Differentiate implicitly

Differentiate $2x^{2}-\sqrt{5}xy + y^{2}=6$ with respect to $x$. Using the power - rule and product - rule: $\frac{d}{dx}(2x^{2})-\sqrt{5}\left(y + x\frac{dy}{dx}\right)+2y\frac{dy}{dx}=0$. $4x-\sqrt{5}y-\sqrt{5}x\frac{dy}{dx}+2y\frac{dy}{dx}=0$.

Step2: Solve for $\frac{dy}{dx}$

Group the terms with $\frac{dy}{dx}$ on one side: $2y\frac{dy}{dx}-\sqrt{5}x\frac{dy}{dx}=\sqrt{5}y - 4x$. $\frac{dy}{dx}(2y-\sqrt{5}x)=\sqrt{5}y - 4x$. So, $\frac{dy}{dx}=\frac{\sqrt{5}y - 4x}{2y-\sqrt{5}x}$.

Step3: Find the slope of the tangent line

Substitute $x = \sqrt{5}$ and $y = 4$ into $\frac{dy}{dx}$: $\frac{dy}{dx}\big|_{x = \sqrt{5},y = 4}=\frac{\sqrt{5}\times4-4\times\sqrt{5}}{2\times4-\sqrt{5}\times\sqrt{5}}=\frac{4\sqrt{5}-4\sqrt{5}}{8 - 5}=0$.

Step4: Find the equation of the tangent line

Use the point - slope form $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(\sqrt{5},4)$ and $m = 0$. $y-4=0(x - \sqrt{5})$. The equation of the tangent line is $y = 4$.

Answer:

$y = 4$