• # question_answer The equation of a common tangent to ${{y}^{2}}=4x$ and the curve ${{x}^{2}}+4{{y}^{2}}=8$can be A) $x-2y+2=0$ B) $x+2y+4=0$C) $x-2y=4$                   D) $x+2y=4$

${{y}^{2}}=4x\And \frac{{{x}^{2}}}{8}+\frac{{{y}^{2}}}{2}=1$ Equation of tangent to above curves are respectively. ${{y}^{2}}=mx+\frac{1}{m}$and$y=mx+\sqrt{8{{m}^{2}}+2}$ Comparing$\frac{1}{m}=\sqrt{8{{m}^{2}}+2}$$\Rightarrow$${{m}^{2}}(8{{m}^{2}}+2)=1$seeing the options $m=\pm \frac{1}{2}$satisfy the equation $\Rightarrow$$y=\pm \frac{1}{2}x\pm 2\Rightarrow 2y=\pm x\pm 4$ $i.e.\,\,2y=x+4\And x+2y+4=0$