Question

Let three distinct normal be drawn to the parabola \(y^{2}+4y-6x-8=0\) from a point \((a, \mathrm{b})\) on the axis of the parabola. Then :

• A. \(a \in (1, \infty)\) and \(b=-2\)
• B. \(a \in (0, \infty)\) and \(b=-2\)
• C. \(a \in (1, \infty)\) and \(b=2\)
• D. \(a \in (2, \infty)\) and \(b=2\)

Hint:

Rewrite the parabola as \((y+2)^2=6(x+2)\); its axis is \(y=-2\). Use the slope form of the normal and demand three distinct real slopes.

Answer 1

The Correct Option is A

Complete the square: \(y^2+4y=6x+8\Rightarrow (y+2)^2=6(x+2)\). Put \(X=x+2,\ Y=y+2\): \(Y^2=6X\), so \(4A=6,\ A=\frac32\). The axis is \(Y=0\), i.e. \(y=-2\). A point on the axis therefore has \(b=-2\) and in new coordinates is \((h,0)\) with \(h=a+2\).
Normal of \(Y^2=4AX\) with slope \(m\): \(Y=mX-2Am-Am^3\), i.e. \(Y=mX-3m-\frac32m^3\).
It passes through \((h,0)\): \(0=mh-3m-\frac32m^3\Rightarrow m\left(h-3-\frac32m^2\right)=0\).
So \(m=0\) or \(m^2=\frac23(h-3)\). Three distinct normals need two further distinct non-zero real slopes, so \(h-3>0\Rightarrow h>3\).
Hence \(a+2>3\Rightarrow a>1\), and \(b=-2\).
Answer: (1) \(a\in(1,\infty)\) and \(b=-2\)