Question

The point of intersection of the lines
(a^3+3)x +ay+a-3 =0 and (a^5+2)x+(a+2)y+2a+3=0 (a real) lies on the y axis for:-
A). No value of a
B). Exactly two values of a
C). More than 2 values of a
D).Exactly one value of a

Answer 1

where are you stuck at?
equate them, put x=0

Answer 2

I found the family of intersection of the two given lines...and then put that x intercept as 0...

Answer 3

(a^3+3)x +ay+a-3 =(a^5+2)x+(a+2)y+2a+3

with x =0 gives
ay+a = ay +2y +2a+3

isolate y

Answer 4

ay+a -3= ay +2y +2a+3**

Answer 5

a=-(2)(y+3)

Answer 6

isolating y will give y intercept :)

Answer 7

A lot ....

Answer 8

Except for one value of a

Answer 9

y = -(a+6)/2

which value?

Answer 10

For a=-6 we will get y intercept 0

Answer 11

thats still on y axis

Answer 12

Yep....

Answer 13

So ans should be for all values of a

Answer 14

yes
which is same as option C

Answer 15

More than 1 value can also be the ans...

Answer 16

Bt no option as that...:)

Answer 17

A,B, D are definitely false.
method of elimination :P