JEE Main 2018 — Parabola Question with Solution

As origin is the only common point to x-axis and y-axis, so origin is the common vertex

Let the equation of two of parabolas be y² = 4ax and x² = 4by

Now latus rectum of both parabolas = 3

$$\therefore$$ 4a = 4b = 3

$$\Rightarrow$$ a = b = $$\frac{3}{4}$$

$$\therefore$$ Two parabolas are y² = 3x and x² = 3y

Suppose y = mx + c is in the common tangent.

$$\therefore$$ y² = 3x $$\Rightarrow$$ (mx + c)² = 3x $$\Rightarrow$$ m²x² + (2mc $$-$$ 3) x + c² = 0

As, the tangent touches at one point only

So, b² $$-$$ 4ac = 0

$$\Rightarrow$$ (2mc $$-$$ 3)² $$-$$ 4m²c² = 0

$$\Rightarrow$$ 4m²c² + 9 $$-$$ 12mc $$-$$ 4m²c² = 0

$$\Rightarrow$$ c = $$\frac{9}{12m}$$ = $$\frac{3}{4m}$$      . . . .(i)

$$\therefore$$ x² = 3y $$\Rightarrow$$ x² = 3 (mx + c ) $$\Rightarrow$$ x² $$-$$ 3mx $$-$$ 3c = 0

Again, b² $$-$$ 4ac = 0

$$\Rightarrow$$ 9m² $$-$$ 4(1) ($$-$$3c) = 0

$$\Rightarrow$$ 9m² = $$-$$ 12c       . . . . .(ii)

From (i) and (ii)

m² = $$\frac{-4c}{3}$$ = $$\frac{-4}{3}$$ $$\left(\frac{3}{4m}\right)$$

$$\Rightarrow$$ m³ = $$-$$ 1 $$\Rightarrow$$ m = $$-$$ 1 $$\Rightarrow$$ c = $$\frac{-3}{4}$$

Hence, y = mx + c = $$-$$ x $$-$$ $$\frac{3}{4}$$

$$\Rightarrow$$ 4 (x + y) + 3 = 0