Three normals are drawn from the point (7, 14) to the parabola `x^2-8x-16 y=0` . Find the coordinates of the feet of the normals.

Explanation

The equation of the given parabola is
`x^(2)-8x-16y=0`
Differentiating w.r.t. x, we get
`2x-8-16(dy)/(dx)=0or(dy)/(dx)=(x-4)/(8)`
Let normal be drawn at point (h,k) on the curve.
`:." "k=(h^(2)-8h)/(16)`
Also, slope of the normal at `(h,k)=(8)/(4-h)`
Therefore, equation of the normal at (h,k) is
`y-kj=((8)/(4-h) )(x-h)`
If the normal passes through (7,14), then we have
`14-(h^(2)-8h)/(16)=((8)/(4-h))(7-h)`
`rArr" "(4-h)(224-h^(2)+8h)=128(7-h)`
`rArr" "h^(3)-12h^(2)-64h=0`
`rArr" "h(h-16)(h+4)=0`
`rArr" "h=0,-4,16`
For h=0, k=0
for h=-4, k=3 and
for h=16, k=8. Therefore, the coordinates of the feet of the normal are (0, 0), (-4,3) and (16,8).