NIMCET 2024 — Mathematics PYQ

Step 1: Write the equation of a normal in slope form

For a standard parabola $y^2=4kx$, the equation of a normal line with slope $m$ is given by the standard formula:

$$y=mx-2km-k{m}^3$$

Let us find the equation of the normal for both given parabolas in terms of the slope $m$.

1. For the second parabola $y^2=4bx$:

   Comparing with the standard form ($k=b$), the equation of its normal is:

   $$y=mx-2bm-b{m}^3$$

   --- (Equation 1)

2. For the first parabola $y^2=4a(x+c)$:

   This is a shifted parabola where $x$ is replaced by $(x+c)$ and $k=a$. Replacing $x$ with $(x+c)$ in the standard normal form gives:

   $$y=m(x+c)-2am-a{m}^3$$

   $$y=mx+mc-2am-a{m}^3$$

   --- (Equation 2)

Step 2: Compare coefficients for a common normal

If these two parabolas have a common normal, then Equation 1 and Equation 2 must represent the exact same line for a non-zero slope $m$.

Comparing the constant y-intercept terms of both equations:

$$-2bm-b{m}^3=mc-2am-a{m}^3$$

Step 3: Simplify the condition for $m$

Bring all the terms involving $m$ to one side:

$$a{m}^3-b{m}^3-2bm+2am-mc=0$$

$$m\left[(a-b)m^2+2(a-b)-c\right]=0$$

This gives two possible scenarios for the slope $m$:

1. $m=0$: This represents the $x$-axis ($y=0$), which is a common axis of symmetry and acts as a trivial common normal for both parabolas.

2. $(a-b)m^2+2(a-b)-c=0$: This gives the condition for a non-trivial, inclined common normal.

Step 4: Find the condition for real existing values of $m$

For an inclined common normal to exist, $m^2$ must yield a real, positive non-zero value (m^2 > 0):

$$(a-b)m^2=c-2(a-b)$$

$$m^2=\frac{c-2(a-b)}{a-b}$$

Since $m^2$ must be strictly greater than $0$ for distinct non-trivial normals, and we are given that a > b > 0 (which means the denominator a - b > 0 is positive):

\frac{c - 2(a - b)}{a - b} > 0

\implies c - 2(a - b) > 0

\implies c > 2(a - b)

Conclusion

The two parabolas cannot have a non-trivial common normal unless c > 2(a - b).

Therefore, the correct option is B.