CircleHard
Question
x - 2y + 4 = 0 is a common tangent to y2 = 4x & 
= 1. Then the value of b and the other common tangent are given by -
= 1. Then the value of b and the other common tangent are given by -Options
A.b = √3 ; x + 2y + 4 = 0
B.b = 3 ; x + 2y + 4 = 0
C.b = √3 ; x + 2y - 4 = 0
D.b = √3 ; x - 2y - 4 = 0
Solution
Equation of tangent of ellipse is
y = mx ±
......... (i)
Given equation is x - 2y + 4 = 0 ......... (ii)
Since (i) & (ii) are same, comparing them, we get
m =
& 
= 2
⇒ 4.
+ b2 = 4
⇒ b = ± √3
Equation of tangent of parabola
y = mx +
......... (iii)
By (i) & (iii)
= a2m2 + b2
on solving it we get m = ±
with m = -
we get x + 2y + 4 = 0
which is other equation of common tangent.
y = mx ±
......... (i)Given equation is x - 2y + 4 = 0 ......... (ii)
Since (i) & (ii) are same, comparing them, we get
m =
& = 2⇒ 4.
+ b2 = 4⇒ b = ± √3
Equation of tangent of parabola
y = mx +
......... (iii)By (i) & (iii)
= a2m2 + b2on solving it we get m = ±
with m = -
we get x + 2y + 4 = 0which is other equation of common tangent.
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