A) 8
B) 4
C) 5
D) 6
Correct Answer: B
Solution :
[b]Let point \[P\,({{t}^{2}},2t)\] lie on parabola \[{{y}^{2}}=4x.\]Image of \[P\,({{t}^{2}},2t)\] with respect to line \[x+y+4=0\]is \[8\,(h,k).\] |
\[\therefore \]\[\frac{h-{{t}^{2}}}{1}=\frac{k-2t}{1}=\frac{-\,2\,({{t}^{2}}+2t+4)}{2}\] |
Hence, curve C becomes |
\[\frac{{{(x+4)}^{2}}}{4}+4=-\,y\] |
Since it intersect with \[y=-\,5\] |
\[\therefore \] \[{{(x+4)}^{2}}+16=20\] |
\[\Rightarrow \]\[{{(x+4)}^{2}}=4\] |
\[\Rightarrow \]\[x=-\,2\]or \[x=6\] |
\[\therefore \]\[A=\,(-2,-\,5)\]and \[B=\,(-\,6,-\,5)\] |
\[\Rightarrow \]\[AB=\sqrt{{{(-2+6)}^{2}}+{{(-\,5+5)}^{2}}}=4\] |
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