Question

A curve is such that `dy/dx=4/sqrt((6-2x))` and `P(1,8)` is a point on the curve. 1. The normal to the curve at the point `P` meets the coordinate axes at `Q` and at `R`. Find the coordinates of the mid-point of `QR.` 2. Find the equation of the curve?

A curve is such that `dy/dx=4/sqrt((6-2x))` and `P(1,8)` is a point on the curve.

1. The normal to the curve at the point `P` meets the coordinate axes at `Q` and at `R`. Find the coordinates of the mid-point of `QR.`
2. Find the equation of the curve?

Answer 1

The coordinates of the midpoint of `QR` is `=(17/2,17/4)`. The equation of the curve is `y=-4(6-2x))^(1/2)+16`

Explanation:

The gradient to the curve `f(x)` is

`dy/dx=f'(x)=4/(sqrt(6-2x))`

At the point `P(1,8)`, the gradient is

`f'(1)=4/sqrt(6-2*1)=4/sqrt4=2`

The slope of the tangent at the point `P` is `m=2`

Therefore,

The slope of the normal at the point `P`

`=m'=-1/m=-1/2`

The equation of the normal at the point `P` is

`y-8=-1/2(x-1)`

`2y-16=-x+1`

`2y+x=17`

When `x=0`, `=>`, `y=17/2`

The point `Q=(0,17/2)`

When `y=0`, `=>`, `x=17`

The point `R=(17,0)`

The midpoint of `QR` is

`=((17+0)/2,(17/2+0)/2)`

`=(17/2, 17/4)`

The equation of the curve is obtained by integrating the derivative, or by solving the differential equation

`y=int(4dx)/(sqrt(6-2x))`

`=4int(6-2x)^(-1/2)`

`=4*-1/2*2/1*(6-2x)^(1/2)+C`

`y=-4(6-2x))^(1/2)+C`

Plugging in the values of `P(1,8)`

`8=-4(6-2)^(1/2)+C`

`C=16`

The equation of the curve is

`y=-4(6-2x))^(1/2)+16`

graph{(2y+x-17)(y+4sqrt(6-2x)-16)=0 [-22.17, 23.44, -4.57, 18.27]}