JEE Main 2019 — Parabola Question with Solution

Normal to the two given curves are

y = m(x – c) – 2bm – bm³,

y = mx – 4am – 2am³

If they have a common normal, then

(c + 2b)m + bm³ = 4am + 2am³

$$\Rightarrow$$ (4a – c – 2b) m = (b – 2a)m³

$$\Rightarrow$$ (4a – c – 2b) = (b – 2a)m²

$$\Rightarrow$$ m² = $$\frac{c}{2a-b}-2$$ $$>$$ 0

By checking all options we found (A) is right option.

Note : We get that all the options are correct for m = 0 i.e., when common normal is x-axis. But may be in question they want common normal other than x – axis, hence answer is (A).