DSE 2011 doubt

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DSE 2011 doubt

Halflife

Q23) Answer : C
I got the answer but does anyone know any quick way to do it?
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Re: DSE 2011 doubt

megha sonik
could you please tell me how we you are getting b<2 cause I'm getting b>2?
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Re: DSE 2011 doubt

Halflife
You've to minimize the distance between (x, x^2/4) and (0, b) where b can take different values greater than 0.
Take b=2 and find the distance. Then take b=3 and find the distance. The distance has increased in case of b=3 so the answer has to be b<2.
But this method took time so I'm asking for any other method that is faster.
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Re: DSE 2011 doubt

megha sonik
I used distance formula and differentiated it wrt to x
but I'm not getting the answer.
could you please do this using that and tell where I'm going wrong?
please!
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Re: DSE 2011 doubt

Halflife
I already tried the distance formula but it got too complicated.
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Re: DSE 2011 doubt

AdityaGarg13
In reply to this post by Halflife
Answer is d)
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Re: DSE 2011 doubt

Halflife
In Amit sir's key it's given as (c) and I have never found a wrong answer in his key in any paper.
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Re: DSE 2011 doubt

Asd1995
In reply to this post by Halflife
Set of all points equidistant from (0,b) is a circle. So set of all points at a distance of b from this point is a circle around it touching the x axis (tangent to it as well). At what b will this circle intersect with the parabola?

Solve x^2+(y-b)^2=b^2 with y=x^2/4

4y+(y-b)^2=b^2

y^2+y(4-2b)=0

y=0, y=2b-4, obviously for the root to make sense it has to have a positive y, since both the circle and parabola lie on y.

So for b<2, closest point is the origin
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Re: DSE 2011 doubt

Halflife
Thanks.
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Re: DSE 2011 doubt

AdityaGarg13
If y= 2b-4, and y has to positive. Then should not be y >2???
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Re: DSE 2011 doubt

Asd1995
We solved for the condition of intersection ie. opposite of what we want.
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