Find Equation of directrix and axis of parabola 4y^2- 6x-4y=5

Answers (1)

The given parabola

                                                          $$ 4y^2 -6x-4y = 5 $$\$$ $$Rightarrow 4(y^2 - y) = 6x + 4 $$\$$ $$Rightarrow (y^2 - y) = frac14(6x +4)$$\$$ $$Rightarrow y^2 - y +frac14 = frac14(6x +5) +frac14$$\$$ $$Rightarrow (y-frac12)^2 = frac14 (6x +6)$$\$$ $$Rightarrow (y-frac12)^2 = frac32 (x +1)$$\$$

we know that the general equation of a parabola 

                                                                      $$ y^2 = 4 a x$$

Here

                    $$ y = y-frac12 and, a = frac38 and, x = x+ 1 $$

Directrix  $$ x = - a$$

                        $$ x +1 = - frac38$$\$$ $$ 8x + 11 = 0$$\$$

The axis

                  $$ y = 0 $$ \$$ $$ y-frac12 = 0$$\$$ $$ 2y -1 = 0 $$ \$$

Hence, Directrics 

                       $$ 8x + 11 $$\$$ and Axis$$ = 2y-1 = 0$$

 

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