Question

Find the equations of the tangent and the normal lines to the given parabola at the given point. Sketch the parabola, the tangent line, and the normal line. $$ x^{2}=-10 y,(2 \sqrt{5},-2) $$

Short answer

The tangent line is \(y + 2 = -\frac{2\sqrt{5}}{5}(x - 2\sqrt{5})\), and the normal line is \(y + 2 = \frac{\sqrt{5}}{2}(x - 2\sqrt{5})\).

Step-by-step solution

Identify the given parabola and point

The equation of the parabola is given as \(x^2 = -10y\). The point of tangency on this parabola is \((2\sqrt{5}, -2)\).

Differentiate the equation to find the slope of the tangent

First, differentiate the equation \(x^2 = -10y\) with respect to \(x\) to find the derivative \(\frac{dy}{dx}\). Differentiating gives \(2x = -10 \frac{dy}{dx}\). Therefore, \(\frac{dy}{dx} = -\frac{x}{5}\).

Calculate the slope of the tangent at the given point

Substitute \(x = 2\sqrt{5}\) into the derivative \(\frac{dy}{dx} = -\frac{x}{5}\) to find the slope at the point of tangency. This gives \(\text{slope of tangent} = -\frac{2\sqrt{5}}{5}\).

Write the equation of the tangent line

Using the point-slope form \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point \((2\sqrt{5}, -2)\), the equation of the tangent line is \(y + 2 = -\frac{2\sqrt{5}}{5}(x - 2\sqrt{5})\).

Calculate the slope of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent. Therefore, the slope is \(\frac{5}{2\sqrt{5}}\). Simplifying, we get \(\text{slope of normal} = \frac{\sqrt{5}}{2}\).

Write the equation of the normal line

Using the point-slope form \(y - y_1 = m(x - x_1)\), with the slope of the normal and point of tangency \((2\sqrt{5}, -2)\), we have: \(y + 2 = \frac{\sqrt{5}}{2}(x - 2\sqrt{5})\).

Sketch the parabola, tangent, and normal lines

To sketch, plot the parabola using \(x^2 = -10y\), which opens downwards. Plot the point \((2\sqrt{5}, -2)\). Then draw the tangent and normal lines using the equations derived: \(y + 2 = -\frac{2\sqrt{5}}{5}(x - 2\sqrt{5})\) and \(y + 2 = \frac{\sqrt{5}}{2}(x - 2\sqrt{5})\). The tangent will touch the parabola at just one point, while the normal will be perpendicular to the tangent.

Key concepts

Understanding Parabolas

Parabolas are U-shaped graphs that represent a quadratic equation, which can be in the form of either \(y = ax^2 + bx + c\) or \(x = ay^2 + by + c\). In this exercise, we deal with a vertical parabola, \(x^2 = -10y\). This means that instead of opening upwards or downwards, it opens to the left or right. Typically, the vertex of a parabola is its point of symmetry, and for this particular equation, the parabola opens downwards because of the negative coefficient.
To plot this specific parabola accurately:

• Begin by identifying the axis of symmetry. Here, it's vertical.
• Note the orientation of the parabola is influenced by the sign of the coefficient of the \(y\) term.
• Plot key points to see the curve's path.

Differentiation and Finding Derivatives

Differentiation is a tool from calculus used to find how a function changes at any given point. When we'd like to know the slope of a curve, we differentiate. Here, the parabola's equation \(x^2 = -10y\) is implicitly defined. Differentiating both sides with respect to \(x\), using implicit differentiation, gives:

• The derivative of \(x^2\) is \(2x\).
• The derivative of \(-10y\) with respect to \(x\) is \(-10 \frac{dy}{dx}\) (since \(y\) is dependent on \(x\)).

This leads to the equation \(2x = -10 \frac{dy}{dx}\), revealing the rate at which \(y\) changes with \(x\). Solving for \(\frac{dy}{dx}\), we find \(\frac{dy}{dx} = -\frac{x}{5}\). This expression tells us the slope of the tangent line anywhere on the parabola.

Calculating Slopes

The slope is a critical measure showing the steepness or the incline of a line. At the given point, \((2\sqrt{5}, -2)\), the slope of the tangent is calculated by substituting into the derivative \(\frac{dy}{dx} = -\frac{x}{5}\). By plugging \(x = 2\sqrt{5}\) into the derivative, we find that the slope of the tangent at the point of tangency is \(-\frac{2\sqrt{5}}{5}\).
The slope is negative, indicating a downward trend. Meanwhile,for the normal line, which is perpendicular to the tangent, the slope is the negative reciprocal of the tangent's slope. Therefore, we have

• For tangent: \(-\frac{2\sqrt{5}}{5}\)
• For normal: \(\frac{5}{2\sqrt{5}}\), simplifying to \(\frac{\sqrt{5}}{2}\)

Formulating the Tangent Equation

The tangent line ‘grazes’ the curve at precisely one point. Its equation is often found using the point-slope formula: \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the given point and \(m\) is the slope.Here, we use the point \((2\sqrt{5}, -2)\) along with the calculated tangent slope: \(-\frac{2\sqrt{5}}{5}\). Placing these values into the equation yields:
\(y + 2 = -\frac{2\sqrt{5}}{5}(x - 2\sqrt{5})\).
This line represents the immediate path that continues from the point of tangency, showing the derivative's influence visually.

Constructing the Normal Equation

To find the normal line, which is perpendicular to the tangent at the given point, we take the tangent'sslope's negative reciprocal. This yields the normal line's slope as \(\frac{\sqrt{5}}{2}\). The normal line alsouses the point-slope equation method with the point \((2\sqrt{5}, -2)\), producing:
\(y + 2 = \frac{\sqrt{5}}{2}(x - 2\sqrt{5})\).
The normal line serves as a perpendicular bisector to some degree,providing geometric balance around the point of tangency on the curve, portraying the curvature's perpendicular response.
To sketch, you would draw:

• The parabola, opening downwards
• The tangent line, barely touching
• The normal line, crossing at a right angle