   Chapter 3.5, Problem 45E

Chapter
Section
Textbook Problem

Find an equation of the tangent line to the hyperbola x 2 a 2 − y 2 b 2 = 1 at the point (x0, y0).

To determine

To find: The equation of the tangent line to the hyperbola at the given point.

Explanation

Given:

The equation of hyperbola is x2a2y2b2=1.

The point is (x0,y0).

Derivative rules: Chain rule

If y=f(u) and u=g(x)  are both differentiable function, then dydx=dydududx.

Formula used:

The equation of the tangent line at (x1,y1) is, yy1=m(xx1) (1)

where, m is the slope of the tangent line at (x1,y1) and m=dydx|x=x1,y=y1

Calculation:

Obtain the equation of the tangent line to the hyperbola at the point (x0,y0).

x2a2y2b2=1

Differentiate the equation implicitly with respect to x,

ddx(x2a2y2b2)=ddx(1)ddx(x2a2)ddx(y2b2)=01a2ddx(x2)1b2ddx(y2)=0

Apply the chain rule and simplify the terms,

1a2(2x)1b2[ddy(y2)dydx]=01a2(2x)1b2[2ydydx]=02xa22yb2dydx=02yb2dydx=2xa2

Multiply the equation by b22y,

dydx=2xa2×b22ydydx=b2xa2y

Thus, the derivative of the equation is dydx=b2xa2y

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