Question

the equation of normal y=5-x² at x=2 is

Answer 1

Answer:

normal=x-4y+2=0

tangent=4x+y-9=0

Step-by-step explanation:

=> i am using differential calculus approach

at y₁=5-x^2=5-4=1 (x₁=2)

so slope(m) of y=5- x^2     dy/dx =-2x = -4 (x₁=2)

so the tangent is

(y-y₁)=m(x-x₁)

so we get

y-1=(-4)*(x-2)

on simplification, we get y+4x-9=0 => tangent

normal is perpendicular to tangent and also meets the curve at (2,1)

{if a₁x+b₁y+c₁=0, then line perpendicular to it would be of the form b₁x-a₁y+k=0}

so applying the perpendicular line condition

we get normal is a line of the form x-4y+c=0

and since it intersects at 2,1

(2) -4(1) +c=0

c=2

hence normal is x-4y+2=0

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