CircleEasy
Question
Two common tngents to the circle x2 + y2 = 2a2 and parabola y2 = 8ax are
Options
A.x = ± (y + 2a)
B.y = ± (x + 2a)
C.x = ± (y + a)
D.y = ± (x + a)
Solution
Any tangent to the parabola y2 = 8ax is
y = mx +
......(i)
If (i) is a tangent to the circle, x2 + y2 = 2a2 then, √2a = ±
⇒ m2 (1 + m2) = 2 ⇒ (m2 + 2) (m2 - 1) = 0 ; ⇒ m = ± 1
So from (i), y = ± (x + 2a)
y = mx +
......(i)If (i) is a tangent to the circle, x2 + y2 = 2a2 then, √2a = ±
⇒ m2 (1 + m2) = 2 ⇒ (m2 + 2) (m2 - 1) = 0 ; ⇒ m = ± 1
So from (i), y = ± (x + 2a)
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