Answer :

The given equation is $y^{2}=12 x$

Here, the coefficient of $x$ is positive.

Hence, the parabola opens towards the right.

On comparing this equation with $y^{2}=4 a x$,

we obtain $4 a=12 $ $\Rightarrow a=3$

$\therefore \ \ $ Coordinates of the focus $=\left(a, 0\right)=\left(3,0\right)$

Since the given equation involves $y^{2}$, the axis of the parabola is the $x$-axis.

Equation of direcctrix, $x=-a \ $ i.e., $x=-3$ i.e., $x+3=0$

Length of latus rectum $=4 a=4 \times 3=12$

Answer :

The given equation is $x^{2}=6 y$

Here, the coefficient of $y$ is positive.

Hence, the parabola opens upwards.

On comparing this equation with $x^{2}=4 a y$,

we obtain $4 a=6 $ $\Rightarrow a=\dfrac{3}{2}$

$\therefore \ \ $ Coordinates of the focus $=\left(0, a\right)=\left(0, \dfrac{3}{2}\right)$

Since the given equation involves $x^{2}$, the axis of the parabola is the $y$-axis.

Equation of directrix,

$ y=-a \text{ i.e., } y=-\dfrac{3}{2} $

Length of latus rectum $=4 a=6$

Answer :

The given equation is $y^{2}=-8 x$

Here, the coefficient of $x$ is negative.

Hence, the parabola opens towards the left.

On comparing this equation with $y^{2}=-4 a x$,

we obtain $-4 a=-8 \Rightarrow a=2$

$\therefore \ \ $ Coordinates of the focus $=\left(-a, 0\right)=\left(-2,0\right)$

Since the given equation involves $y^{2}$, the axis of the parabola is the $x$-axis.

Equation of directrix, $x=$ a $\quad$ i.e., $x=2$

Length of latus rectum $=4 a=8$

Answer :

The given equation is $x^{2}=-16 y$

Here, the coefficient of $y$ is negative.

Hence, the parabola opens downwards.

On comparing this equation with $x^{2}=-4 a y$,

we obtain $-4 a=-16 \Rightarrow a=4$

$\therefore \ \ $ Coordinates of the focus $=\left(0,-a\right)=\left(0,-4\right)$

Since the given equation involves $x^{2}$, the axis of the parabola is the $y$-axis.

Equation of directrix, $y=a\quad$ i.e., $y=4$

Length of latus rectum $=4 a=16$

Answer :

The given equation is $y^{2}=10 x$

Here, the coefficient of $x$ is positive.

Hence, the parabola opens towards the right.

On comparing this equation with $y^{2}=4 a x$,

we obtain

$4 a=10 \Rightarrow a=\dfrac{5}{2}$

$\therefore \ \ $ Coordinates of the focus $=\left(a, 0\right)$ $ =\left(\dfrac{5}{2}, 0\right) $

Since the given equation involves $y^{2}$, the axis of the parabola is the $x$-axis.

Equation of directrix,

$ x=-a \quad\text{ i.e., } x=-\dfrac{5}{2} $

Length of latus rectum $=4 a=10$

Answer :

The given equation is $x^{2}=-9 y$

Here, the coefficient of $y$ is negative.

Hence, the parabola opens downwards.

On comparing this equation with $x^{2}=4a y $

we obtain $-4 a=-9 \Rightarrow b=\dfrac{9}{4}$

$\therefore \ \ $ Coordinates of the focus $=\left(0,-a\right)=\left(0,-\dfrac{9}{4}\right)$

Since the given equation involves $x^{2}$, the axis of the parabola is the $y$-axis.

Equation of directrix,

$ y=a \quad\text{ i.e., } y=\dfrac{9}{4} $

Length of latus rectum $=4 a=9$

Answer :

Focus $\left(6,0\right)$; directrix, $x=-6$

Since the focus lies on the $x$-axis, the $x$-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form $y^{2}=4 a x$ or $y^{2}=-4 a x$

It is also seen that the directrix, $x=-6$ is to the left of the $y$-axis, while the focus $\left(6,0\right)$ is to the right of the $y$-axis.

Hence, the parabola is of the form $y^{2}=4 a x$

Here, $a=6$

Thus, the equation of the parabola is $y^{2}=24 x$

Answer :

Vertex $\left(0,0\right)$; focus $\left(3,0\right)$

Since the vertex of the parabola is $\left(0,0\right)$ and the focus lies on the positive $x$-axis, $x$-axis is the axis of the parabola, while the equation of the parabola is of the form $y^{2}=4 a x$

Since the focus is $\left(3,0\right), a=3$

Thus, the equation of the parabola is $y^{2}=4x \times 3 \quad $, i.e., $y^{2}=12 x$

Answer :

Focus $=\left(0,-3\right) ;$ directrix $y=3$

Sincethe focus lies on the $y$-axis, the $y$-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form $x^{2}=4ay$ or $x^{2}=-4 a y$

It is also seen that the directrix, $y=3$ is above the $x$-axis, while the focus $\left(0,-3\right)$ is below the $x$-axis.

Hence, the parabola is of the form $x^{2}=-4 a y$

Here, $a=3$

Thus, the equation of the parabola is $x^{2}=-12 y$

Answer :

Vertex $\left(0,0\right)$ focus $\left(-2,0\right)$

Since the vertex of the parabola is $\left(0,0\right)$ and the focus lies on the negative $x$-axis, $x$-axis is the axis of the parabola, while the equation of the parabola is of the form $y^{2}=-4 a x$

Since the focus is $\left(-2,0\right), a=2$

Thus, the equation of the parabola is $y^{2}=-4\left(2\right) x\quad$,i.e., $y^{2}=-8 x$

Answer :

Since the vertex is $\left(0,0\right)$ and the axis of the parabola isthe $x$-axis, the equation of the parabola is either of the form $ ~ y^{2}=4 a x ~ $ or $ ~ y^{2}=- 4 a x$

The parabola passes through point $\left(2,3\right)$, which lies in the first quadrant.

Therefore, the equation of the parabola is of the form $y^{2}=4 a x$, while point $\left(2,3\right)$ must satisfy the equation $y^{2}=4 a x$

$\therefore \ \ 3^{2}=4 a\left(2\right) \Rightarrow a=\dfrac{9}{8}$

Thus, the equation of the parabola is

$ \begin{aligned} & y^{2}=4\left(\dfrac{9}{8}\right) x \\ \\ & y^{2}=\dfrac{9}{2} x \\ \\ & 2 y^{2}=9 x \end{aligned} $

Answer :

Since the vertex is $\left(0,0\right)$ and the parabola is symmetric aboutthe $y$-axis, the equation of the parabola is either of the form $ ~ x^{2}=4 ay$ or $ ~ x^{2}=- 4ay$

The parabola passes through point $\left(5,2\right)$, which lies in the first quadrant.

Therefore, the equation of the parabola is of the form $x^{2}=4 a y$, while point $\left(5,2\right)$ must satisfy the equation $x^{2}=4 a y$

$\therefore \ \ \left(5\right)^{2}=4 \times a \times 2 $

$\quad\Rightarrow 25=8 a \Rightarrow a=\dfrac{25}{8}$

Thus, the equation of the parabola is $x^{2}=4\left(\dfrac{25}{8}\right) y$

$2 x^{2}=25 y$