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

Explanation

The given parabola is
`y^(2)-16x-8y=0` ….(i)
Let the co-ordinates of the feet of the normal from (14, 7) be `P(alpha,beta)`. Now the equation of the tangent at `P(alpha,beta)` to parabola (i) is
`ybeta-8(x+alpha)-4(y+beta)=0`
or `(beta-4)y=8x+8a+4beta` ...(ii)
Its slope
`=8/(beta-4)`
Equation of the normal to parabola (i) at `(alpha,beta)` is y `– beta`
`=(4-beta)/8(x-alpha)`
It passes through (14, 7)
`!7-beta=(4-beta)/8(14-alpha)`
`!alpha=(6beta)/(beta-4)` ...(iii)
Also `(alpha,beta)` lies on parabola (i)
i.e., `beta^(2)-16alpha-8beta=0` ...(iv)
Putting the value of  from (iii) in (iv), we get
`beta^(2)-(96beta)/(beta-4)-8beta=0`
`!beta^(2)(beta-4)-96beta-8beta(beta-4)=0`
`!beta(beta^(2)-4beta-96-8beta+32)=0`
`!beta(beta^(2)-12beta-64)=0`
`!(beta-16)(beta+4)=0`
`!beta=0,16,-4`
from (iii), `alpha` = 0 when `beta = 0, alpha = 8,` when `beta = 16 , alpha = 3` when `beta = –4`
Hence the feet of the normals are (0, 0) (8, 16) and (3, –4)