The given circle #x^2 + y^2 + 2px = 0#, #p inRR# touches the parabola #y^2 = 4x# externally, then( prove this mathematically)?

Question

A)#p<0#
B)#p>0#
C)#0<p<1#
D)#p<-1#

The answer is B) for your reference.

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Answer 1 · 2018-01-31T06:44:39

Answer is (B) #p>0#

The parabola #y^2=4x# is in #Q1# and #Q2# and passes through #(0,0)#.

Further, equation of circle #x^2+y^2+2px=0# can be written as #(x+p)^2+y^2=p^2# and hence center of circle is #(-p,0)# and radius is #p#.

Therefore, the circle will touch the parabola externally if centre is

#p>0#

If #p<0#, circle touches internally as centre is #(-p,0)#

Below we show three circles touching parabola externally if #p=0.5,2,4#. One circle with #p=-2#, which touches internally is also shown.

graph{(x^2+y^2+x)(x^2+y^2+8x)(x^2+y^2+4x)(x^2+y^2-4x)(y^2-4x)=0 [-9.83, 10.17, -4.965, 5.035]}

Hence, answer is (B).