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Equation of a line which is tangent t...

Equation of a line which is tangent to both the curve `y=x^2+1\ a n d\ y=x^2` is `y=sqrt(2)x+1/2` (b) `y=sqrt(2)x-1/2` `y=-sqrt(2)x+1/2` (d) `y=-sqrt(2)x-1/2`

A

`y=sqrt2x-(1)/(2)`

B

`y=sqrt2x+(1)/(2)`

C

`y=-sqrt2x+(1)/(2)`

D

`y=-sqrt2x-(1)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
B, C, D
Let the tangent line be `y=ax+b`
The equation for its intersection with the upper parabola is
`x^(2)+1=ax+b" "rArr" "x^(2)-ax(1-b)=0`
This has equal roots when `a^(2)-4(1-b)=0 or a^(2)+4b=4(1)`
For the lower parabola
`ax+b=-x^(2)rArr x^(2)+ax+b=09`
This has equal roots when `a^(2)-4b=0" (2)"`
From (1) and (2) `8b=4 or b=1//4`
`"Add "2a^(2)=4" or "a=pm sqrt2`
The tangent lines are `y=sqrt2x+(1)/(2) and y=-sqrt2x+(1)/(2)`
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