39–50. Equations of ellipses and hyperbolas Find an equation of the following ellipses and hyperbolas, assuming the center is at the origin.
Table of contents
- 0. Functions7h 54m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms36m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
- 12. Techniques of Integration7h 41m
- 13. Intro to Differential Equations2h 55m
- 14. Sequences & Series5h 36m
- 15. Power Series2h 19m
- 16. Parametric Equations & Polar Coordinates7h 58m
16. Parametric Equations & Polar Coordinates
Conic Sections
Problem 12.4.69
Textbook Question
69–72. Tangent lines Find an equation of the line tangent to the following curves at the given point.
x² = -6y; (-6, -6)
Verified step by step guidance1
Rewrite the given curve equation \(x^{2} = -6y\) to express \(y\) explicitly in terms of \(x\). This will help in differentiating \(y\) with respect to \(x\).
Differentiate both sides of the equation \(x^{2} = -6y\) implicitly with respect to \(x\). Remember to apply the chain rule when differentiating \(y\) since \(y\) is a function of \(x\).
After differentiating, solve for \(\frac{dy}{dx}\), which represents the slope of the tangent line at any point on the curve.
Substitute the given point \((-6, -6)\) into the expression for \(\frac{dy}{dx}\) to find the slope of the tangent line at that specific point.
Use the point-slope form of a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the slope found in the previous step and \((x_1, y_1)\) is the given point, to write the equation of the tangent line.
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Key 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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Slope of the Tangent Line
The slope of the tangent line at a given point on a curve is the value of the derivative dy/dx evaluated at that point. It represents the instantaneous rate of change of y with respect to x and determines the steepness and direction of the tangent line.
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Slopes of Tangent Lines
Equation of a Tangent Line
Once the slope of the tangent line and the point of tangency are known, the equation of the tangent line can be found using the point-slope form: y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point on the curve.
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