Question

In Exercises 17-26, find the lines that are (a) tangent and (b) normal to the curve at the given point. $$x^{2}-\sqrt{3} x y+2 y^{2}=5, \quad(\sqrt{3}, 2)$$

Short answer

The equations for the tangent and normal lines to the curve \(x^{2}-\sqrt{3} x y+2 y^{2}=5\) at the point \((\sqrt{3}, 2)\) are respectively, \(y = 2x - 2\sqrt{3}\) and \(y = -1/2x + 2 + \sqrt{3}/2\).

Step-by-step solution

Find the Implicit Derivative of the Curve

Given the curve \(x^{2}-\sqrt{3} x y+2 y^{2}=5\), differentiate it implicitly with respect to x. This entails applying the chain rule to \(y^{2}\) by treating \(y\) as a function of \(x\), and the product rule to \(-\sqrt{3} xy\). Be mindful that when differentiating a product of two factors, the derivative of the product is the derivative of the first factor times the second factor plus the first factor times the derivative of the second factor. After taking the derivative, simplify it.

Evaluation of the Derivative at the Given Point

Evaluate the derivative at the given point \((\sqrt{3}, 2)\). This results in the slope of the tangent line at that point.

Formulate the Equation for the Tangent Line

With the slope of the tangent line and the coordinates of a point on the line, calculate the equation for this tangent line using the point slope form \(y - y1 = m(x - x1)\)

Find the Slope of the Normal Line

As the normal line is perpendicular to the tangent line, its slope is the negative reciprocal of the slope of the tangent line.

Formulate the Equation for the Normal Line

The slope of the normal line and the coordinates of a point on the line which is the given point are used to construct the equation for this normal line.