Question

What is the equation of the normal line to the graph of `y=2xsqrt(x^2+8) + 2`?

Answer 1

Thus, the equation of the norma is given by

`y=3/2xsqrt(x^2+8)+2`

Explanation:

Given
`y=2xsqrt(x^2+8)+2`
At any point on the graph, the normal has slope perpendicular to the slope of the tangent at the point given by the first derivative of the function.

`(dy)/dx=2xxx1/(2sqrt(x^2+8))xx2x+0=(2x^2)/sqrt(x^2+8)`

Slope of the tangent `m=(2x^2)/sqrt(x^2+8)`

Thus the normal has the slope equal to the negative reciprocal

Slope of the normal `m'=(-sqrt(x^2+8))/2`

Intercept made by the straight line on y axis is given by
`c=y-mx=y-((-sqrt(x^2+8))/2x)`

Substituting for `y` and simplifying

`c=(2xsqrt(x^2+8)+2)+(xsqrt(x^2+8))/2`
`=(2x+x/2)sqrt(x^2+8)+2=(5x)/2sqrt(x^2+8)+2`

`c=(5x)/2sqrt(x^2+8)+2`

Equation of a straight line havihg slope m and intercept as c is given by

`y=mx+c`

`y=(-sqrt(x^2+8))/2x+(5x)/2sqrt(x^2+8)+2`

`=(-1+5/2)xsqrt(x^2+8)+2`

`=3/2xsqrt(x^2+8)+2`

Thus, the equation of the normal is given by

`y=3/2xsqrt(x^2+8)+2`