Question

Find the vertex, focus, axis, directrix and latus-rectum of the following parabolas
(i) y² = 8x
(ii) 4x² + y = 0
(iii) y² − 4y − 3x + 1 = 0
(iv) y² − 4y + 4x = 0
(v) y² + 4x + 4y − 3 = 0
(vi) y² = 8x + 8y
(vii) 4 (y − 1)² = − 7 (x − 3)
(viii) y² = 5x − 4y − 9
(ix) x² + y = 6x − 14

Answer 1

(i) Given:
y² = 8x

On comparing the given equation with $y^2=4ax$:
$4a=8\Rightarrow a=2$

∴ Vertex = (0, 0)

Focus = (a, 0) = (2, 0)

Equation of the directrix:
x = −a
i.e. x = −2

Axis = y = 0

Length of the latus rectum = 4a = 8 units

(ii) Given:
4x² + y = 0

$\Rightarrow \frac{-y}{4}=x^2$

On comparing the given equation with $x^2=-4ay$:

$4a=\frac{1}{4}\Rightarrow a=\frac{1}{16}$

∴ Vertex = (0, 0)

Focus = (0, −a) = $\left(0,\frac{-1}{16}\right)$

Equation of the directrix:
y = a
i.e. $y=\frac{1}{16}$

Axis = x = 0

Length of the latus rectum = 4a = $\frac{1}{4}$ units

(iii) Given:
y² − 4y − 3x + 1 = 0

$$\begin{gathered} \Rightarrow {\left(y-2\right)}^2-4-3x+1=0 \\ \Rightarrow {\left(y-2\right)}^2=3\left(x+1\right) \\ \Rightarrow {\left(y-2\right)}^2=3\left(x-\left(-1\right)\right) \end{gathered}$$

Let $Y=y-2$, $X=x+1$
Then, we have:
$Y^2=3X$

Comparing the given equation with $Y^2=4aX$:

$4a=3\Rightarrow a=\frac{3}{4}$

∴ Vertex = (X = 0, Y = 0) = $\left(x=-1,y=2\right)$

Focus = (X = a, Y = 0) = $\left(x+1=\frac{3}{4},y-2=0\right)=\left(x=\frac{-1}{4},y=2\right)$

Equation of the directrix:
X = −a
i.e. $x+1=\frac{-3}{4}\Rightarrow x=\frac{-7}{4}$

Axis = Y = 0
i.e. $y-2=0\Rightarrow y=2$

Length of the latus rectum = 4a = 3 units

(iv) Given:
y² − 4y + 4x = 0

$$\begin{gathered} \Rightarrow {\left(y-2\right)}^2-4+4x=0 \\ \Rightarrow {\left(y-2\right)}^2=-4\left(x-1\right) \end{gathered}$$

Let $Y=y-2$, $X=x-1$
Then, we have:
$Y^2=-4X$

Comparing the given equation with $Y^2=-4aX$:

$4a=4\Rightarrow a=1$

∴ Vertex = (X = 0, Y = 0) = $\left(x=1,y=2\right)$

Focus = (X = −a, Y = 0) = $\left(x-1=-1,y-2=0\right)=\left(x=0,y=2\right)$

Equation of the directrix:
X = a
i.e. $x-1=1\Rightarrow x=2$

Axis = Y = 0
i.e. $y-2=0\Rightarrow y=2$

Length of the latus rectum = 4a = 4 units

(v) Given:
y² + 4y + 4x −3 = 0

$$\begin{gathered} \Rightarrow {\left(y+2\right)}^2-4+4x-3=0 \\ \Rightarrow {\left(y+2\right)}^2=-4\left(x-\frac{7}{4}\right) \end{gathered}$$

Let $Y=y+2$, $X=x-\frac{7}{4}$
Then, we have:
$Y^2=-4X$

Comparing the given equation with $Y^2=-4aX$:
$4a=4\Rightarrow a=1$

∴ Vertex = (X = 0, Y = 0) = $\left(x=\frac{7}{4},y=-2\right)$

Focus = (X = −a, Y = 0) = $\left(x-\frac{7}{4}=-1,y+2=0\right)=\left(x=\frac{3}{4},y=-2\right)$

Equation of the directrix:
X = a
i.e. $x-\frac{7}{4}=1\Rightarrow x=\frac{11}{4}$

Axis = Y = 0
i.e. $y+2=0\Rightarrow y=-2$

Length of the latus rectum = 4a = 4 units

(vi) Given:
y² = 8x + 8y
$\Rightarrow {\left(y-4\right)}^2=8\left(x+2\right)$

Putting $Y=y-4$, $X=x+2$:
$Y^2=8X$


On comparing the given equation with $Y^2=4aX$:
$4a=8\Rightarrow a=2$

∴ Vertex = (X = 0, Y = 0) = $\left(x=-2,y=4\right)$

Focus = (X = a, Y = 0) = $\left(x+2=2,y-4=0\right)=\left(x=0,y=4\right)$
Equation of the directrix:
X = −a
i.e. $x+2=-2\Rightarrow x+4=0$
Axis = Y = 0
i.e. $y-4=0\Rightarrow y=4$
Length of the latus rectum = 4a = 8

(vii) Given:
4(y − 1)² = − 7 (x − 3)
$\Rightarrow {\left(y-1\right)}^2=\frac{-7}{4}\left(x-3\right)$

Let $Y=y-1$, $X=x-3$
Then, we have:
$Y^2=\frac{-7}{4}X$

Comparing the given equation with $Y^2=-4aX$:
$4a=\frac{7}{4}\Rightarrow a=\frac{7}{16}$

∴ Vertex = (X = 0, Y = 0) = $\left(x=3,y=1\right)$

Focus = (X = −a, Y = 0) = $\left(x-3=\frac{-7}{16},y-1=0\right)=\left(x=\frac{41}{16},y=1\right)$

Equation of the directrix:
X = a
i.e. $x-3=\frac{7}{16}\Rightarrow x=\frac{55}{16}$

Axis = Y = 0
i.e. $y-1=0\Rightarrow y=1$

Length of the latus rectum = 4a = $\frac{7}{4}$ units

(viii) Given:
y ² = 5x − 4y − 9

$$\begin{gathered} \Rightarrow y^2+4y=5x-9 \\ \Rightarrow {\left(y+2\right)}^2=5x-5=5\left(x-1\right) \end{gathered}$$

Putting $Y=y+2$, $X=x-1$:
$Y^2=5X$

Comparing the given equation with $Y^2=4aX$:
$4a=5\Rightarrow a=\frac{5}{4}$

∴ Vertex = (X = 0, Y = 0) = $\left(x=1,y=-2\right)$

Focus = (X = a, Y = 0) = $\left(x-1=\frac{5}{4},y+2=0\right)=\left(x=\frac{9}{4},y=-2\right)$

Equation of the directrix:
X = −a
i.e. $x-1=\frac{-5}{4}\Rightarrow x=\frac{-1}{4}$

Axis = Y = 0
i.e. $y+2=0\Rightarrow y=-2$

Length of the latus rectum = 4a = 5 units

(ix) Given:
x² = 6x−y−14
$$\begin{gathered} \Rightarrow {\left(x-3\right)}^2=-y-14+9 \\ \Rightarrow {\left(x-3\right)}^2=-y-5=-\left(y+5\right) \end{gathered}$$

Let $Y=y+5$, $X=x-3$
Then, we have:
$X^2=-Y$

Comparing the given equation with $X^2=-4aY$:
$4a=1\Rightarrow a=\frac{1}{4}$

∴ Vertex = (X = 0, Y = 0) = $\left(x=3,y=-5\right)$

Focus = (X = 0, Y = −a) = $\left(x-3=0,y+5=\frac{-1}{4}\right)=\left(x=3,y=\frac{-21}{4}\right)$

Equation of the directrix:
Y = a

i.e. $y+5=\frac{1}{4}\Rightarrow y=\frac{-19}{4}$

Axis = X = 0
i.e. $x-3=0\Rightarrow x=3$

Length of the latus rectum = 4a = 1 units