Question

Three normals are drawn from the point $(7,14)$ to the parabola $x^2-8x-16y=0$. Then the coordinates of the foot of the normal is/are

• A. $(0,0)$
• B. $(4,-1)$
• C. $(-4,3)$
• D. $(16,8)$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $(0,0)$
C $(-4,3)$
D $(16,8)$

Equation of the parabola,
$x^2-8x-16y=0\Rightarrow (x-4{)}^2=16y+16\Rightarrow (x-4{)}^2=16(y+1)$
General equation of normal (in slope form) is
$(x-4)=m(y+1)-2\times 4m-4m^3$
Putting the point $(7,14)$ in the equation of the normal,
$3=15m-8m-4m^3\Rightarrow 4m^3-7m+3=0$
$m=1$ is a root of the equation,
$(m-1)(4m^2+4m-3)=0$
$(4m^2+4m-3)=0\Rightarrow 4m^2+6m-2m-3=0\Rightarrow (2m+3)(2m-1)=0\Rightarrow m=-\frac{3}{2}\text{or}\frac{1}{2}$
Coordinates of the foot will be
$x-4=-2am=-8m\Rightarrow x=4-8m\Rightarrow x=-4,16,0$
$y+1=a{m}^2=4m^2\Rightarrow y=4m^2-1\Rightarrow y=3,8,0$

Hence the coordinates will be $(0,0)$ $(-4,3)$ and $(16,8)$