Question

Determine the equation of the normal line to y = x–2/x at (2, 0)​

Answer 1

Answer:

The given curve is

y=2x−x2

i.e, x2−2x+y=0

Now the equation of the tangent at the point (x1,y1)=(2,0)

or, xx1−(x+x1)+21(y+y1)=0

i.e 2x+(−1)(x+2)+21(y+0)=0

i.e,2x−4+y=0

i.e,y=−2x+4

So, the slope of the tangent is (−2)

Therefore the slope of the normal=21

So,the equation of the normal at the point (2,0) is

y−0=21(x−2)

i.e, 2y=x−2

i.e, x−2y=2

Step-by-step explanation:

I hope it's helpful to!!

Answer 2

Answer:

Here is your answer

Step-by-step explanation:

The given curve is

   y=2x−x2

i.e, x2−2x+y=0

Now the equation of the tangent at the point (x1,y1)=(2,0)

or, xx1−(x+x1)+21(y+y1)=0

i.e 2x+(−1)(x+2)+21(y+0)=0

i.e,2x−4+y=0

i.e,y=−2x+4

So, the slope of the tangent is (−2)

Therefore the slope of the normal=21

So,the equation of the normal at the point (2,0) is

y−0=21(x−2)

i.e, 2y=x−2

i.e, x−2y=2

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