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CalculusCalculus10th Edition

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Chapter 10.1, Problem 49E
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Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095

Solutions

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Problem 1E
Problem 2E
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Problem 5E
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Problem 7E
Problem 8E
Problem 9E
Problem 10E
Problem 11E
Problem 12E
Problem 13E
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Problem 15E
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Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
Problem 26E
Problem 27E
Problem 28E
Problem 29E
Problem 30E
Problem 31E
Problem 32E
Problem 33E
Problem 34E
Problem 35E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
Problem 41E
Problem 42E
Problem 43E
Problem 44E
Problem 45E
Problem 46E
Problem 47E
Problem 48E
Problem 49E
Problem 50E
Problem 53E
Problem 51E
Problem 52E
Problem 57E
Problem 58E
Problem 55E
Problem 54E
Problem 63E
Problem 69E
Problem 56E
Problem 59E
Problem 60E
Problem 61E
Problem 62E
Problem 64E
Problem 65E
Problem 66E
Problem 67E
Problem 68E
Problem 70E
Problem 71E
Problem 72E
Problem 73E
Problem 74E
Problem 75E
Problem 76E
Problem 77E
Problem 78E
Problem 79E
Problem 80E
Problem 81E
Problem 82E
Problem 83E
Problem 84E
Problem 85E
Problem 86E
Problem 87E
Problem 88E
Problem 89E
Problem 90E
Problem 91E
Problem 92E
Problem 93E
Problem 94E
Problem 95E
Problem 96E
Problem 97E
Problem 98E
BuyFindarrow_forward

Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095
Textbook Problem

Finding Equations of Tangent Lines and Normal Lines In Exercises 49 and 50, find equations fur (a) the tangent lines and (b) the normal lines to the hyperbola for the given value of x. (The normal line at a point is perpendicular to the tangent line at the point.)

x 2 9 − y 2 = 1 ,   x = 6

(a)

To determine

To-calculate: The tangent lines for the hyperbola x29−y2=1 at x=6.

Explanation

Given:

The equation of the hyperbola, x29−y2=1 and the point x=6.

Formula used:

Equation of a line passing through (x1,y1) and slope m is (y−y1)=m(x−x1).

Calculation:

Consider the provided equation of the hyperbola, x29−y2=1.

Now, substitute x=6 in the equation of the hyperbola, x29−y2=1, to find the coordinate for y.

So,,

369−y2=1−y2=1−369−y2=273y=±3

thus, the two coordinates are (6,3) and (6,−3).

Now, differentiate the equation of the hyperbola, x29−y2=1 with respect to x.

Then,

2x9−2ydydx=0.

dydx=x9y …...…... (1)

Therefore, dydx=x9y is the slope of the curve at any point on the curve of the hyperbola, x29−y2=1

So, the slope of the hyperbola, x29−y2=1 at (6,3) will be,

dydx=69(3)=233=239

Now, consider the standard equation of the line (y−y1)=m(x−x1) to calculate the equation of the tangent passing through (6,3) and with a slope 239

(b)

To determine

To-calculate: The equations of the normal lines for the hyperbola x29−y2=1 at x=6.

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Chapter 10.1,
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