(d) : Statement 1 :
\(y^2=16\sqrt3x,\)
Therefore, the equation of tangent to the given parabola is \(y=mx+\frac{4\sqrt3}{m}\) ......(1)
\(\frac{x^2}{2}+\frac{y^2}{4}=1,\)
Therefore, the equation of tangent to the given ellipse is \(x=m_1y+\sqrt{4m_1^2+2}\)
\(\Rightarrow y=\frac{x}{m_1}-\sqrt{4+\frac{2}{m_1^2}}\) .....(2)
Since, equations (1) and (2) represents common tangents to both ellipse and parabola.
Therefore, by comparing equations (1) and (2), we obtain
\(m=\frac{1}{m_1}\)
And
\((\frac{4\sqrt3}{m})^2=\bigg(-\sqrt{4+\frac{2}{m_1^2}}\bigg)^2\)
\(\Rightarrow \frac{48}{m^2}=4+\frac{2}{m_1^2}\)\(=4+2m^2\) (\(\because\, m=\frac{1}{m_1}\))
\(\Rightarrow\frac{24}{m^2}=2+m^2\)
\(\Rightarrow m^4+2m^2-24=0...(3)\)
\(\Rightarrow(m^2+6)(m^2-4)=0\)
\(\Rightarrow m=\pm2\)
Statement 2:
If \(y=mx+\frac{4\sqrt3}{m}\) is a common tangent
to \(y^2=16\sqrt3x\) and ellipse \(2x^2+y^2=4,\) then m
satisfies \(m^4+2m^2-24=0\). (From equation (3))
Hence, statement (2) is true.
Now, by putting m=2 in equation (1), we get the common tangent which is
\(y=2x+\frac{4\sqrt3}{2}\)
\(\Rightarrow y=2x+{2\sqrt3}\).
Hence, statement (1) is true.
Thus, statement (2) is a correct explanation for statement (1).
