Textbook Question
53–57. Conic sections
a. Determine whether the following equations describe a parabola, an ellipse, or a hyperbola.
x²/4 + y²/25 = 1
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16. Parametric Equations & Polar Coordinates
Conic Sections
2:47 minutesProblem 12.R.58
Textbook Question
58–59. Tangent lines Find an equation of the line tangent to the following curves at the given point. Check your work with a graphing utility.
x²/16 - y²/9 = 1; (20/3, -4)
Verified step by step guidance1
Identify the given curve: \(\frac{x^2}{16} - \frac{y^2}{9} = 1\) and the point of tangency \(\left(\frac{20}{3}, -4\right)\).
Differentiate both sides of the equation implicitly with respect to \(x\). Remember to apply the chain rule when differentiating terms involving \(y\), treating \(y\) as a function of \(x\). The differentiation will look like this: \(\frac{d}{dx}\left(\frac{x^2}{16}\right) - \frac{d}{dx}\left(\frac{y^2}{9}\right) = \frac{d}{dx}(1)\).
Calculate each derivative: \(\frac{d}{dx}\left(\frac{x^2}{16}\right) = \frac{2x}{16} = \frac{x}{8}\), and for \(\frac{d}{dx}\left(\frac{y^2}{9}\right)\) use the chain rule: \(\frac{2y}{9} \cdot \frac{dy}{dx}\). Set up the equation: \(\frac{x}{8} - \frac{2y}{9} \cdot \frac{dy}{dx} = 0\).
Solve for \(\frac{dy}{dx}\) (the slope of the tangent line) by isolating it: \(\frac{dy}{dx} = \frac{\frac{x}{8}}{\frac{2y}{9}} = \frac{x}{8} \cdot \frac{9}{2y} = \frac{9x}{16y}\).
Substitute the coordinates of the given point \(\left(\frac{20}{3}, -4\right)\) into the expression for \(\frac{dy}{dx}\) to find the slope at that point. Then use the point-slope form of a line: \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point, to write the equation of the tangent line.
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Video duration:
2mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Implicit Differentiation
Implicit differentiation is used when a function is given in an implicit form, such as an equation involving both x and y. Instead of solving for y explicitly, we differentiate both sides with respect to x, treating y as a function of x, and apply the chain rule to find dy/dx.
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Finding The Implicit Derivative
Equation of a Tangent Line
The equation of a tangent line at a point on a curve is given by y - y₁ = m(x - x₁), where m is the slope of the tangent line (dy/dx at the point) and (x₁, y₁) is the point of tangency. Finding the slope is essential to write this linear equation.
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Hyperbola and Its Properties
The given curve is a hyperbola defined by x²/16 - y²/9 = 1. Understanding its shape and symmetry helps in visualizing the curve and verifying the tangent line. The hyperbola has two branches and asymptotes, which influence the behavior of tangent lines.
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