Question

if the point P(4a,2a-1) lies on the graph of the equation x-2y=2 then the number of values (s) of is are​

Answer 1

Step-by-step explanation:

Point(2a,a+1)isaninteriorpointofcircle,

∴(2a)

2

+(a+1)

2

−2(2a)−2(a+1)−8<0.

⇒4a

2

+a

2

+1+2a−4a−2a−2−8<0.

⇒5a

2

−4a−9<0.

⇒5a

2

−9a+5a−9<0.

⇒5a

2

−5a−9a−9<0.

⇒5a(a+1)−9(a+1)<0.

⇒(a+1)(a−

5

9

)<0.

a ε (−1,

5

9

).→(1).

Point will lie on same side of line towards which centre of circle lie.

⇒Centre→(1,1).

⇒x−y+1

⇒1−1+1=2>0.

∴2a−(a+1)+1>0.

⇒2a−a−1+1>0.

⇒a>0→(2).

Using(1)&(2):

⇒aε(0,

5

9

)

Answer 2

Step-by-step explanation:

since the point lies on graph of equation x-2y=2, so the point P Satisfies it.. so put values of x and y in given equation

$4a - 2(2a - 1) = 2 \\ 4a - 4a + 2 = 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2 = 2$

so clearly on putting values of x and y equation reduces to identity or a vanishes.... hence there exist infinitely many values of a