   Chapter 10.1, Problem 49E

Chapter
Section
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 x29y2=1 at x=6.

Explanation

Given:

The equation of the hyperbola, x29y2=1 and the point x=6.

Formula used:

Equation of a line passing through (x1,y1) and slope m is (yy1)=m(xx1).

Calculation:

Consider the provided equation of the hyperbola, x29y2=1.

Now, substitute x=6 in the equation of the hyperbola, x29y2=1, to find the coordinate for y.

So,,

369y2=1y2=1369y2=273y=±3

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

Now, differentiate the equation of the hyperbola, x29y2=1 with respect to x.

Then,

2x92ydydx=0.

dydx=x9y …...…... (1)

Therefore, dydx=x9y is the slope of the curve at any point on the curve of the hyperbola, x29y2=1

So, the slope of the hyperbola, x29y2=1 at (6,3) will be,

dydx=69(3)=233=239

Now, consider the standard equation of the line (yy1)=m(xx1) 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 x29y2=1 at x=6.

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