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consider the common point of parabola and circle.
since y2=4ax
circle=x2+y2+2kx=O
so, x2+4ax +kx=O
now from condition of tangency,
(4a+2k)2+O=O
or,16a2+4k2+16ak=O
since a is real
so,(16k)2+4.4k2.16>O
or,k>O
The circle x2+y2+2kx=0,k belongs to R,touches the parabola y2=4x externally.Then
(a)k>0 (b)k<0 (c)k>1 (d)none of these
Ans is (a).
i have put y2=4x in the eq.of the circle.and in the quad.of x,i put b2=4ac.i got k=-2.where am i wrong?
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3 Answers
661in your method you have not considered any condition for the circle to touch the parabola externally.
so, according to your answer k=-2, the circle is touching the parabola internally but we need to find the value of k for which the circle touches the parabola externally.
solution:-
the equation of all the points lying outside the parabola y2=4x is
y2>4x...(i)
centre of the given circle is (-2k,0)
as the circle touches the parabola externally,
its centre lies outside the parabola.
putting x=-2k and y=0 in (i) we get,
02>4(-2k)
-8k<0
8k>0
k>0
- Debargha Paul Centre of the circle is not (-2k,0).It is (-k,0).Upvote·0· Reply ·2013-04-02 07:33:12
- Gaurav Gardi yes the centre of the circle is (-k,0).
then in my solution instead of 8k>0 it will be 4k>0 which also implies that k>0.
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