Two parabolas with a common vertex and with axes along the x-axis and y-axis, respectively, intersect each other in the first quadrant. If the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is?
Solutions · 1
Official 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 =43\therefore\,\,\,\[ Two parabolas are y2 = 3x and x2 = 3y
Suppose y = mx + c is in the common tangent.
\]\therefore\,\,\,y2 = 3x⇒(mx + c)2 = 3x⇒m2x2 + (2mc-\[ 3) x + c2 = 0
As, the tangent touches at one point only
So, b2 \]-4ac = 0⇒(2mc−3)2−4m2c2 = 0⇒4m2c2 + 9−12mc−4m2c2 = 0⇒c =12m9=4m3. . . .(i)∴x2 = 3y⇒x2 = 3 (mx + c )⇒x2−3mx−3c = 0
Again, b2−4ac = 0⇒9m2−4(1) (−3c) = 0⇒9m2 =-\[ 12c . . . . .(ii)
From (i) and (ii)
m2 = \]{{ - 4c} \over 3}=3−4\left( {{3 \over {4m}}} \right) (
\] \Rightarrow \[ \,\,\,\] m3 = - 1 \[ \Rightarrow \] m = - 1 \[ \Rightarrow \] c = \[{{ - 3} \over 4}\]
Hence, y = mx + c = - x - \[{3 \over 4}\]
\[ \Rightarrow \] 4 (x + y) + 3 = 0