Question

The equation of the axis of the parabola 2x² + 5y – 3x + 4 = 0 is __________.

Answer 1

Given parabola is,
$$\begin{gathered} 2x^2+5y-3x+4=0 \\ \mathrm{i}.\mathrm{e}.2x^2-3x=-5y-4 \\ \mathrm{i}.\mathrm{e}.x^2-\frac{3x}{2}=\frac{-5y}{2}-\frac{4}{2} \\ \mathrm{i}.\mathrm{e}.x^2-2x\left(\frac{3}{4}\right)=\frac{-5y}{2}-\frac{4}{2} \\ \mathrm{i}.\mathrm{e}.x^2-2x\left(\frac{3}{4}\right)+{\left(\frac{3}{4}\right)}^2-{\left(\frac{3}{4}\right)}^2=\frac{-5y}{2}-\frac{4}{2} \\ \mathrm{i}.\mathrm{e}.{\left(x-\frac{3}{4}\right)}^2-{\left(\frac{3}{4}\right)}^2=\frac{-5y}{2}-\frac{4}{2} \\ \mathrm{i}.\mathrm{e}.{\left(x-\frac{3}{4}\right)}^2=\frac{-5}{2}y-\frac{4}{2}+\frac{9}{16} \\ \mathrm{i}.\mathrm{e}.{\left(x-\frac{3}{4}\right)}^2=\frac{-5}{2}y-2+\frac{9}{16} \\ \mathrm{i}.\mathrm{e}.{\left(x-\frac{3}{4}\right)}^2=\frac{-5}{2}y-\frac{23}{16} \\ \mathrm{i}.\mathrm{e}.{\left(x-\frac{3}{4}\right)}^2=\frac{-5}{2}\left(y+\frac{23}{40}\right) \\ \therefore \mathrm{axis}\mathrm{of}\mathrm{parabola}\mathrm{is}\mathrm{given}\mathrm{by}x-\frac{3}{4}=0 \\ \mathrm{i}.\mathrm{e}.x=\frac{3}{4} \end{gathered}$$