The set of points on the axis of the parabola ${{y}^{2}}-4x-2y+5=0$ form which all the three normals to the parabola are real is:
[a] $\left\{ \left( x,1 \right):x\ge 3 \right\}$
[b] $\left\{ \left( x,-1 \right):x\ge 1 \right\}$
[c] $\left\{ \left( x,3 \right):x\ge 1 \right\}$
[d] $\left\{ \left( x,-3 \right):x\ge 3 \right\}$
VerifiedHint: Shift the origin to (1,1) and find the condition for a point to have three normals passing through it. Use the property that a cubic $f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ has all three roots real and distinct if and only if $3a{{x}^{2}}+2bx+c=0$ has real and distinct roots ${{x}_{1}}$ and ${{x}_{2}}$ and $f\left( {{x}_{1}} \right)\ne 0,f\left( {{x}_{2}} \right)\ne 0$. Revert to the original coordinate system and hence find the condition so that three normals pass through that point.
Complete step-by-step solution -
We have ${{y}^{2}}-4x-2y+5=0$
Hence we have ${{\left( y-1 \right)}^{2}}=4\left( x-1 \right)$
Shifting the origin to (1,1) and let the new coordinate system be X-Y.
Hence we have X = x-1 and Y = y-1
Hence the equation of the parabola in the new coordinate system is ${{Y}^{2}}=4X$.
Equation of normal parabola ${{y}^{2}}=4ax$ is given by
$y=mx-2am-a{{m}^{3}}$.
Hence the equation of the normal to the parabola ${{Y}^{2}}=4X$ is given by
$Y=mX-2m+{{m}^{3}}$
Since this line passes through the axis of the parabola at (a,0) [say], we have
$ 0=ma-2m-{{m}^{3}} $
$ \Rightarrow {{m}^{3}}+m\left( 2-a \right)=0 $
$ \Rightarrow m\left( {{m}^{2}}+2-a \right)=0 $
Since three normals pass through (a,0), we have three real values of m that satisfy the above equation.
One real value is m = 0.
Hence the quadratic ${{m}^{2}}+a-2$=0 has real roots.
Hence we have
$ a-2\ge 0 $
$ \Rightarrow a\ge 2 $
Hence we have
$ x-1\ge 2 $
$ \Rightarrow x\ge 3 $
Hence the set points on the x-axis from which all the three normal to the parabola are real is
$\left\{ \left( x,1 \right):x\ge 3 \right\}$
Hence option [a] is correct.
Note: [1] Alternatively, we have if from a point (h,k) all the normals to the parabola ${{y}^{2}}=4ax$ are real, then $h\ge 2a$
Here a = 1 and h = x-1
Hence we have $x-1\ge 2\Rightarrow x\ge 3$
Hence option [a] is correct.
[2] The feet of normals which pass through a common point are known as co-normal points.
