Question

The equation of the directrix of the parabola x^{2}-2x-y-3=0 is​

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation of parabola is

$\rm \: {x}^{2} - 2x - y - 3 = 0$

can be rewritten as

$\rm \: {x}^{2} - 2x = y + 3$

On adding 1 on both sides, we get

$\rm \: {x}^{2} - 2x + 1 = y + 3 + 1$

$\rm \: {(x - 1)}^{2} = y + 4$

which can be rewritten as

$\rm \: {X}^{2} = Y$

where,

$\rm \: \: \: \: \: \: X = x - 1$

$\rm \: \: \: \: \: \: Y = y + 4$

Now, on comparing with equation of parabola

$\rm \: {X}^{2} = 4aY$

we get

$\rm \: 4a = 1$

$\rm\implies \:a = \dfrac{1}{4}$

So,

$\rm \: Equation \: of \: directrix \: is\: given \: by$

$\rm \: Y = \: - \: a$

$\rm \: y + 4 = \: - \: \dfrac{1}{4}$

$\rm \: 4y + 16 = \: - \: 1$

$\rm\implies \:\rm \: 4y + 17 = 0$

$\rule{190pt}{2pt}$

Additional Information :-

General equation of conic is

$\begin{gathered}\rm \: {ax}^{2} + 2hxy + {by}^{2} + 2gx + 2fy + c = 0 \\ \end{gathered}$

Nature of Conic

$\begin{gathered} \\ \begin{gathered}\boxed{\begin{array}{c|c} \bf Condition & \bf Nature \: of \: conic \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \triangle \ne \: 0, \: ab - {h}^{2} > 0 & \sf Ellipse \\ \\ \sf \triangle \ne \: 0, \: ab - {h}^{2} < 0 & \sf Hyperbola \\ \\ \sf \triangle \ne \: 0, \: ab - {h}^{2} > 0, \: a + b = 0 & \sf Hyperbola(rectangular)\\ \\ \sf \triangle \ne \: 0, \: h = 0, \: a = b & \sf Circle \end{array}} \\ \end{gathered} \\ \end{gathered}$