Question

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\\}

Answer 1

Hint: 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.