Question

If the parabolas y² = 4b(x – c) and y² = 8ax have a common normal, then which one of the following is a valid choice for the ordered triad (a, b, c)?
1. \(\left( {\frac{1}{2},2,{\rm{\;}}3} \right)\)
2. (1, 1, 3)
3. \(\left( {\frac{1}{2},2,\;0} \right)\)
4. (1, 1, 0)

Answer 1

Correct Answer - Option 2 : (1, 1, 3)

The normal to the parabola y² = 4ax in terms of slope is given by the equation:

y = mx – 2am - am³

Now, normal to parabola y² = 4b(x – c) is given as:

⇒ y = m(x – c) – 2bm – bm³

∴ y = mx – (2b + c)m – bm³

Now, normal to parabola y² = 8ax is given as:

∴ y = mx – 4am – 2am³

For common normal, it is given as:

⇒ mx – 4am – 2am³ = mx – (2b + c)m – bm³

⇒ 4am + 2am³ = (2b + c)m + bm³

⇒ (2a – b)m³ + (4a – 2b – c)m = 0

⇒ m((2a – b)m² + (4a – 2b – c)) = 0

∴ m = 0

If m = 0, then all options satisfy because y = 0 is a common normal.

Or

\(\Rightarrow {m^2} = \frac{{2b + c - 4a}}{{2a - b}}\) 

\(\Rightarrow {m^2} = \frac{{c - 2\left( {2a - b} \right)}}{{2a - b}}\) 

\(\Rightarrow {m^2} = \frac{{\left( {2a - b} \right)\left[ {\frac{c}{{\left( {2a - b} \right)}} - 2} \right]}}{{2a - b}}\) 

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

As m² > 0, then,

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

If common normal is other than the axis, then only option (b) satisfies.

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

\(\Rightarrow \frac{3}{{2 - 1}} > 2\) 

∴ 3 > 2