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Ch.12 - Parametric and Polar Curves
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh.12 - Parametric and Polar CurvesProblem 12.R.72a
Chapter 12, Problem 12.R.72a

Parabola-hyperbola tangency: Let P be the parabola y = px² and H be the right half of the hyperbola x² - y² = 1.
a. For what value of p is P tangent to H?

Verified step by step guidance
1
Start by writing down the equations of the two curves: the parabola \(P\) is given by \(y = p x^{2}\), and the hyperbola \(H\) is given by \(x^{2} - y^{2} = 1\) (considering only the right half, so \(x \geq 0\)).
Substitute the expression for \(y\) from the parabola into the hyperbola equation to find the points of intersection. This gives: \(x^{2} - (p x^{2})^{2} = 1\), which simplifies to \(x^{2} - p^{2} x^{4} = 1\).
Rewrite the equation as a polynomial in \(x^{2}\): \(p^{2} x^{4} - x^{2} + 1 = 0\). Let \(z = x^{2}\) to get \(p^{2} z^{2} - z + 1 = 0\).
For the parabola and hyperbola to be tangent, the equation must have exactly one solution for \(z\), meaning the quadratic in \(z\) has a single root. Set the discriminant of this quadratic to zero: \(\Delta = (-1)^{2} - 4 \cdot p^{2} \cdot 1 = 0\).
Solve the discriminant equation \(1 - 4 p^{2} = 0\) for \(p\) to find the value(s) of \(p\) where the parabola is tangent to the hyperbola.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equation of Tangency Between Curves

Tangency between two curves occurs when they touch at a point and share the same slope there. This means their equations must be satisfied simultaneously, and their derivatives (slopes) must be equal at the point of contact.
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Finding Area Between Curves on a Given Interval

Parametric and Implicit Differentiation

To find slopes of curves defined implicitly or explicitly, differentiation techniques are used. For the hyperbola given implicitly, implicit differentiation helps find dy/dx, while for the parabola given explicitly, direct differentiation applies.
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Differentiation of Parametric Curves

Substitution Method for Solving Systems

Solving for tangency involves substituting the parabola's equation into the hyperbola's equation to find common points. This reduces the problem to solving an equation in terms of p, enabling determination of the parameter for tangency.
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Euler's Method
Related Practice
Textbook Question

7–8. Parametric curves and tangent lines

a. Eliminate the parameter to obtain an equation in x and y.

x = 4sin 2t, y = 3cos 2t, for 0 ≤ t ≤ π; t = π/6

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10–12. Parametric curves

a. Eliminate the parameter to obtain an equation in x and y.

x = t² + 4, y = -t, for -2 < t < 0; (5, 1)

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Textbook Question

53–57. Conic sections

b. Use analytical methods to determine the location of the foci, vertices, and directrices.

4x² + 8y² = 16

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Textbook Question

Cartesian conversion Write the equation x=y ² in polar coordinates and state values of θ that produce the entire graph of the parabola.

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Textbook Question

14–18. Parametric descriptions Write parametric equations for the following curves. Solutions are not unique.

The segment of the curve x=y ³ +y+1 that starts at (1, 0) and ends at (11, 2).

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Textbook Question

Eliminate the parameter in the parametric equations x=1+sin t, y=3+2 sin t, for 0≤t≤π/2, and describe the curve, indicating its positive orientation. How does this curve differ from the curve x=1+sin t, y=3+2 sin t, for π/2≤t≤π?

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