Add both the equations to get rid of y..
\((3p – x)^2 – 1 + y^2 +9x^2 – y^2 = 0....9p^2-6px+x^2+9x^2-1=0.....10x^2-6px+(9p^2-1)=0\)
Now for this to have a unique negative solution, it should be able to be written in the form \((x+a)^2=x^2+2ax+a^2\)
So the value of p has to be NEGATIVE to make -6px positive...ONLY D possible..

If you have time and you want to solve it further
\(10x^2-6px+(9p^2-1)=x^2-\frac{6px}{10}+\frac{9p^2-1}{10}=x^2+2ax+a^2\)
Thus \(\frac{-6px}{10}=2ax......a=\frac{-3p}{10}\)
Also \(\frac{9p^2-1}{10}=a^2=(\frac{3p}{10})^2=\frac{9p^2}{100}.......90p^2-10=9p^2.......81p^2=10....9p=\sqrt{10}..or -\sqrt{10}....p=\frac{-\sqrt{10}}{9}\) as p is negative

D
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