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EEJ MAIN Mathematics QUESTION #6043
Question 1

Given a circle $2x^2+2y^2=5$ and a parabola $y^2=4\sqrt{5}\,x$, consider:
Statement I: A common tangent is $y=x+\sqrt{5}$.
Statement II: If $y=mx+\frac{\sqrt{5}}{m}$ is a common tangent, then $m^4-3m^2+2=0$.

  • Both statements true; II explains I✔️
  • Both statements true; II does not explain I
  • Statement I true; Statement II false
  • Statement I false; Statement II true
Correct Answer Explanation

For the parabola $y^2=4\sqrt{5}x$: tangent in slope form is $y=mx+\frac{\sqrt{5}}{m}$.

For the circle $x^2+y^2=5/2$ (radius $\sqrt{5/2}$): distance from origin to tangent $= \frac{|\sqrt{5}/m|}{\sqrt{1+m^2}}=\sqrt{5/2}$

$\frac{5/m^2}{1+m^2}=\frac{5}{2}\Rightarrow\frac{1}{m^2(1+m^2)}=\frac{1}{2}\Rightarrow m^2+m^4=2\Rightarrow m^4+m^2-2=0$

Hmm: $m^4+m^2-2=(m^2+2)(m^2-1)=0$, so $m=\pm1$. Statement II gives $m^4-3m^2+2=0=(m^2-1)(m^2-2)=0$, so $m=\pm1,\pm\sqrt{2}$ — this is different. But $m=\pm1$ works for both. At $m=1$: tangent is $y=x+\sqrt{5}$ ✓. Statement I is true; Statement II is false (wrong equation).