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If y=2x-3 is tangent to the parabola y^2...

If `y=2x-3` is tangent to the parabola `y^2=4a(x-1/3),` then `a` is equal to `(22)/3` (b) `-1` (c) `(14)/3` (d) `(-14)/3`

A

`(22)/(3)`

B

`-1`

C

`(14)/(3)`

D

`(14)/(3)`

Text Solution

Verified by Experts

The correct Answer is:
D
(4) Solving y=2x-3
and `y^(2)=4a(x-(1)/(3))`
Eliminating y, we get
`(2x-3)^(2)=4a(x-(1)/(3))`
`or4x^(2)+9-12x=4ax-(4a)/(3)`
`or4x^(2)-4(3+a)x+9+(4a)/(3)=0`
This equation must have equal roots, i.e.,
D=0
`or16(3+a)^(2)-16(9+(4a)/(3))=0`
`or9+a^(2)+6a=9+(4a)/(3)`
`ora^(2)+(14a)/(3)=0`
`ora=0ora=-(14)/(3)`
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