Question

Question #65cb3

Answer 1

The co-ordinates of the other point `(-7/6, -217/36)` or `(-1.167, -6.028)`

Explanation:

Given -

`y=4x-x^2`

Co-ordinate of the normal line`(5, -5)`

Slope of the parabola at any point is given by its first derivative.

`dy/dx=4-2x`

Slope of the parabola at `x=5`

`dy/dx=4-2(5)=4-10=-6`

Slope of the normal is given by -

`m_2=-1xx1/(m_1)`

[This is derived from the principle - If two lines cut each other verticaly then the products of their slopes is equal `-1`; i.e., `m_1 xx m_2 = -1`]
in our case `m_1=-6` [slope of the parabola at x = 5]

Then the slope of the line cuting it verticaly is `m_2`

We have calculate it using the relation `m_2=-1xx1/(m_1)`

`m_2=-1 xx 1/(-6)=1/6`

The slope of the normal is `m_2=1/6`

We have to find the equation of the normal.

`m_2x+c=y`

Where

`m_2=1/6`
`x=5`
`y=-5`

`1/6 xx5+c=-5`
`5/6+c=-5`
`c=-5-5/6=(-30-5)/6=(-35)/6`
The equation of the normal is
`y=1/6x-35/6`

This line cuts the parabola at `(5, -5)` and at another point.
To find the other point, we have to equate the two equations

`4x-x^2=1/6x -35/6`

`-x^2+4x-1/6 x=-35/6`

`-x^2+(24-1)/6 x=-35/6`

`-x^2+23/6 x=-35/6`

Divide both sides by `-1`

`x^2-23/6 x=35/6`

`x^2-23/6 x+529/144=35/6 + 529/144=(840+529)/144=1369/144`

`(x-23/12)^2=1369/144`

`(x-23/12)=+-sqrt(1369/144)=+-37/12`

`x=37/12+23/12=60/12=5`

`x=-37/12+23/12=(-37+23)/12=(-14)/12=-7/6`

At `x=-7/6`

`y=4x-x^2`
`y=4(-7/6)-(-7/6)^2`
`y=-28/6-49/36=(-168-49)/36=-217/36`

The co-ordinates of the other point `(-7/6, -217/36)` or `(-1.167, -6.028)`