Parabola

BITSAT Online › Co-ordinate Geometry questions › Parabola

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?

#Mathematics #Co-ordinate Geometry

UP 0 DOWN 0 0 3
Post on FacebookPost anonymously?

3 Answers

661

Gaurav Gardi ·2013-04-02 02:01:09

in 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

UP 0 DOWN 2
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.
Upvote·0· Reply ·2013-04-02 23:42:37
Post on FacebookPost anonymously?

Debargha Paul ·2013-04-02 07:45:58

solution-
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