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91Ó°ÊÓ

In Problems 92–98, use Descartes’s method from Problem 91 to find the equation of the line tangent to each graph at the given point.

94.x2+y=5; at(-2,1).

Short Answer

Expert verified

The equation of the tangent line is4x-y+9=0.

Step by step solution

01

Step 1. Given information

The general form of the equation of the tangent line must be y=mx+b.

Substitute (-2,1)in y=mx+b.

1=m(-2)+b1+2m=bb=2m+1

Substitute b=2m+1in y=mx+b.

y=mx+(2m+1)

So the system of equations here is

y=mx+(2m+1)x2+y=5.

02

Step 2.  Solve the system of equations.

Substitute y=mx+(2m+1)in x2+y=5.

role="math" localid="1646976038805" x2+mx+(2m+1)=5x2+mx+(2m-4)=0

Using the formula for quadratic equation,

role="math" x=-m±m2-4(1)(2m-4)2(1)=-m±m2-4(2m-4)2

For a unique solution, the two roots must be equal which means

m2-4(2m-4)=0m2-8m+16=0(m-4)2=0 m-4=0m=4

03

Step 3. Substitute m=4 in y=mx+b.

We get

y=4x+(2×4+1)y=4x+94x-y+9=0

So the equation of the tangent line forx2+y=5at(-2,1)is4x-y+9=0.

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