The equation of the line that passes through `(10 ,-1)` and is perpendicular to `y=(x^2)/4-2` is `4x+y=39` (b) `2x+y=19` `x+y=9` (d) `x+2y=8`

Options

A. 4x+y=39
B. 2x+y=19
C. x+y=9
D. x+2y=8

Answer

D

Explanation

(4) `4y=x^(2)=8`
`or" "4(dy)/(dx)=2x`

`:." Slope of normal "=-(2)/(x_(1))`
But
Slope of normal `=(y_(1)+1)/(x_(1)-10)`
`:." "(y_(1)+1)/(x_(1)-10)=-(2)/(x_(1))`
`orx_(1)y_(1)+x_(1)=-2x_(1)+20`
`orx_(1)y_(1)+3x_(1)=20`
Substituting
`y_(1)=(x_(1)^(2)-8)/(4)` (From the given equation)
we get `x_(1)((x_(1)^(2)-8)/(4)+3)=20`
`orx_(1)(x_(1)^(2)+4)=80`
`orx_(1)^(2)+4x_(1)-80=0`
which has one root, `x_(1)=4`.
Hence, `x_(1)=4anyy_(1)=2`.
Therefore, `P-=(4,2)`.
Therefore, the equation of PA is
`y+1=-(1)/(2)(x-10)`
`or2y+2=-x+10`
`orx+2y-8=0`