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 $=(a, 0)=(3,0)$

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 $=(0, a)=(0, \dfrac{3}{2})$

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 $=(-a, 0)=(-2,0)$

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

Equation of directrix, $x=$ a 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 $=(0,-a)=(0,-4)$

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

Equation of directrix, $y=a$ 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 $=(a, 0)$

$ =(\dfrac{5}{2}, 0) $

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

Equation of directrix,

$ x=-a \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

$-4 a=-9 \Rightarrow a=\dfrac{9}{4}$

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

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{9}{4} $

Length of latus rectum $=4 a=9$

Answer :

Focus $(6,0)$; 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 $(6,0)$ 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 $(0,0)$; focus $(3,0)$

Since the vertex of the parabola is $(0,0)$ 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 $(3,0), a=3$.

Thus, the equation of the parabola is $y^{2}=4 \times 3 x $,

i.e., $y^{2}=12 x$

Answer :

Focus $=(0,-3) ;$ 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 $x^{2}=4$ ay or $x^{2}=-4 a y$.

It is also seen that the directrix, $y=3$ is above the $x$-axis, while the focus

$(0,-3)$ 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 $(0,0)$ focus $(-2,0)$

Since the vertex of the parabola is $(0,0)$ 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 $(-2,0), a=2$.

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

Answer :

Since the vertex is $(0,0)$ 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 $(2,3)$, which lies in the first quadrant.

Therefore, the equation of the parabola is of the form $y^{2}=4 a x$, while point

$(2,3)$ must satisfy the equation $y^{2}=4 a x$.

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

Thus, the equation of the parabola is

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

Answer :

Since the vertex is $(0,0)$ 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 $(5,2)$, which lies in the first quadrant.

Therefore, the equation of the parabola is of the form $x^{2}=4 a y$, while point

$(5,2)$ must satisfy the equation $x^{2}=4 a y$.

$\therefore(5)^{2}=4 \times a \times 2 \Rightarrow 25=8 a \Rightarrow a=\dfrac{25}{8}$

Thus, the equation of the parabola is

$x^{2}=4(\dfrac{25}{8}) y$

$2 x^{2}=25 y$