Q1. \(y^2 = 12x\)

Solution

Given the equation of the parabola is \[y^{2}=12x\] To compare it with the standard form of a parabola opening towards the positive \(x\)-axis, we rewrite it as \[y^{2}=4ax\] On comparing, we have \[4a=12\] which gives \[a=3\]

For a parabola of the form \[y^{2}=4ax\] the focus lies at the point \[(a,0)\] Substituting the value of \(a\), the coordinates of the focus are therefore \[(3,0)\]

The directrix of a parabola represented by \[y^{2}=4ax\] is the line \[x=-a\] Hence, the equation of the directrix is \[x=-3\]

The axis of the parabola is the line about which the curve is symmetric. For the standard equation \(y^{2}=4ax\), the axis coincides with the \(x\)-axis, whose equation is \(y=0\).

The length of the latus rectum of a parabola \[y^{2}=4ax\] is given by \(4a\). Substituting \(a=3\), the length of the latus rectum is \[4\times 3=12\]

Q2. \(x^2 = 6y\)

Solution

The given equation of the parabola is \(x^{2}=6y\). To bring it into the standard form of a parabola whose axis is along the positive \(y\)-direction, we compare it with \(x^{2}=4ay\). Writing the given equation suitably, we have

\[ \begin{aligned} x^{2}&=6y\\ x^{2}&=4\cdot \dfrac{3}{2}\,y\\ x^{2}&=4ay\\ a&=\dfrac{3}{2} \end{aligned} \]

For a parabola of the form \(x^{2}=4ay\), the focus lies on the axis at a distance \(a\) from the vertex. Therefore, the coordinates of the focus are \[\left(0,\dfrac{3}{2}\right)\]

The axis of this parabola is the line about which it is symmetric. Since the equation involves \(x^{2}\) and opens upward along the \(y\)-axis, the axis of the parabola is the \(y\)-axis.

The directrix of a parabola represented by \(x^{2}=4ay\) is the line \(y=-a\). Substituting the value of \(a\), the equation of the directrix is \[y=-\dfrac{3}{2}\]

The length of the latus rectum of a parabola \(x^{2}=4ay\) is given by \(4a\). Hence, its length is

\[ \begin{aligned} 4a&=4\times \dfrac{3}{2}\\ &=6 \end{aligned} \]

Q3. \(x^2 = 6y\)

Solution

The given equation of the parabola is \(y^{2}=-8x\). To compare it with the standard form of a parabola whose axis is along the \(x\)-axis, we write it in the form \(y^{2}=4ax\). Accordingly, we have

\[ \begin{aligned} y^{2}&=-8x\\ y^{2}&=4\cdot(-2)x\\ y^{2}&=4ax\\ a&=-2 \end{aligned} \]

For a parabola of the form \(y^{2}=4ax\), the focus is located at the point \((a,0)\) on the axis of the parabola. Substituting the value of \(a\), the coordinates of the focus are \[(-2,0)\]

The axis of the parabola is the line along which it is symmetric. Since the parabola opens towards the negative direction of the \(x\)-axis, its axis coincides with the \(x\)-axis.

The directrix of a parabola represented by \(y^{2}=4ax\) is the line \(x=-a\). Using the value of \(a=-2\), the equation of the directrix is therefore \[x=2\]

The length of the latus rectum of a parabola is given by the numerical value of \(4a\). Hence, the length of the latus rectum is

\[ \begin{aligned} 4a&=4\times 2\\ &=8 \end{aligned} \]

Q4. \(x^2 = – 16y\)

Solution

The given equation of the parabola is \(x^{2}=-16y\). To compare it with the standard form of a parabola whose axis is along the \(y\)-axis, we write it in the form \(x^{2}=4ay\). On comparison, we obtain

\[ \begin{aligned} x^{2}&=-16y\\ x^{2}&=4\cdot(-4)y\\ a&=-4 \end{aligned} \]

For a parabola represented by \(x^{2}=4ay\), the focus lies on the axis at a distance \(a\) from the vertex. Substituting the value of \(a\), the coordinates of the focus are \[(0,-4)\]

The axis of the parabola is the line about which the curve is symmetric. Since the parabola opens downward along the negative direction of the \(y\)-axis, its axis coincides with the \(y\)-axis.

The directrix of a parabola of the form \(x^{2}=4ay\) is the line \(y=-a\). Hence, the equation of the directrix is \[y=4\]

The length of the latus rectum of a parabola is given by the numerical value of \(4a\). Therefore, the length of the latus rectum is

\[ \begin{aligned} 4a&=4\times 4\\ &=16 \end{aligned} \]

Q5. \(y^2 = 10x\)

Solution

The given equation of the parabola is \(y^{2}=10x\). To compare it with the standard form of a parabola opening along the positive \(x\)-axis, we rewrite it in the form \(y^{2}=4ax\). Thus, we have

\[ \begin{aligned} y^{2}&=10x\\ y^{2}&=4\cdot \dfrac{5}{2}x\\ y^{2}&=4ax\\ a&=\dfrac{5}{2} \end{aligned} \]

For a parabola of the form \(y^{2}=4ax\), the focus lies on the \(x\)-axis at a distance \(a\) from the vertex. Hence, the coordinates of the focus are \[\left(\dfrac{5}{2},0\right)\]

The axis of the parabola is the line along which the curve is symmetric. Since the parabola opens to the right along the \(x\)-axis, its axis coincides with the \(x\)-axis.

The directrix of a parabola represented by \(y^{2}=4ax\) is the line \(x=-a\). Substituting the value of \(a\), the equation of the directrix is \[x=-\dfrac{5}{2}\]

The length of the latus rectum of a parabola is given by \(4a\). Therefore, the length of the latus rectum is

\[ \begin{aligned} 4a&=4\cdot \dfrac{5}{2}\\ &=10 \end{aligned} \]

Q6. \(x^2 = – 9y\)

Solution

The given equation of the parabola is \(x^{2}=-9y\). To compare it with the standard form of a parabola whose axis is along the \(y\)-axis, we write it in the form \(x^{2}=4ay\). On comparison, we obtain

\[ \begin{aligned} x^{2}&=-9y\\ x^{2}&=4\cdot\left(-\dfrac{9}{4}\right)y\\ x^{2}&=4ay\\ a&=-\dfrac{9}{4} \end{aligned} \]

For a parabola represented by \(x^{2}=4ay\), the focus lies on the axis at a distance \(a\) from the vertex. Substituting the value of \(a\), the coordinates of the focus are \[\left(0,-\dfrac{9}{4}\right)\]

The axis of the parabola is the line about which the curve is symmetric. Since the parabola opens downward along the negative direction of the \(y\)-axis, its axis coincides with the \(y\)-axis.

The directrix of a parabola of the form \(x^{2}=4ay\) is the line \(y=-a\). Hence, the equation of the directrix is \[y=\dfrac{9}{4}\]

The length of the latus rectum of a parabola is given by the numerical value of \(4a\). Therefore, the length of the latus rectum is

\[ \begin{aligned} 4a&=4\times \dfrac{9}{4}\\ &=9 \end{aligned} \]

Q7. Focus (6,0); directrix \(x = – 6\)

Solution

The given focus of the parabola is \((6,0)\) and the equation of the directrix is \(x=-6\). Since the focus and the directrix are symmetric about the origin and lie along the \(x\)-axis, the axis of the parabola is the \(x\)-axis and the parabola opens towards the positive \(x\)-direction.

For a parabola whose axis is the \(x\)-axis and whose vertex is at the origin, the standard equation is \(y^{2}=4ax\), where the focus is \((a,0)\) and the directrix is \(x=-a\).

Comparing the given focus \((6,0)\) and directrix \(x=-6\) with the standard form, we obtain \(a=6\). Substituting this value of \(a\) in the standard equation, we get

\[ \begin{aligned} y^{2}&=4ax\\ y^{2}&=4\times 6x\\ y^{2}&=24x \end{aligned} \]

Hence, the required equation of the parabola is \(y^{2}=24x\).

Q8. Focus (0,–3); directrix \(y = 3\)

Solution

The given focus of the parabola is \((0,-3)\) and the equation of the directrix is \(y=3\). Since the focus lies below the origin on the \(y\)-axis and the directrix lies above the origin, the axis of the parabola is the \(y\)-axis and the parabola opens downward.

For a parabola whose axis is the \(y\)-axis and whose vertex is at the origin, the standard equation is \(x^{2}=4ay\), where the focus is \((0,a)\) and the directrix is \(y=-a\).

Comparing the given focus \((0,-3)\) and the directrix \(y=3\) with the standard form, we obtain \(a=-3\). Substituting this value of \(a\) in the standard equation, we get

\[ \begin{aligned} a&=-3\\ x^{2}&=4ay\\ x^{2}&=4\cdot(-3)y\\ x^{2}&=-12y \end{aligned} \]

Hence, the required equation of the parabola is \(x^{2}=-12y\).

Q9. Vertex (0,0); focus (3,0)

Solution

The vertex of the parabola is given as \((0,0)\) and the focus is \((3,0)\). Since the focus lies on the positive \(x\)-axis, the axis of the parabola is the \(x\)-axis and the parabola opens towards the positive direction of the \(x\)-axis.

For a parabola whose vertex is at the origin and whose axis is the \(x\)-axis, the standard equation is \(y^{2}=4ax\), where the focus is \((a,0)\).

Comparing the given focus \((3,0)\) with the standard form, we obtain \(a=3\). Substituting this value of \(a\) in the standard equation, we get

\[ \begin{aligned} y^{2}&=4ax\\ a&=3\\ y^{2}&=4\cdot 3x\\ y^{2}&=12x \end{aligned} \]

Hence, the required equation of the parabola is \(y^{2}=12x\).

Q10. Vertex (0,0); focus (–2,0)

Solution

The vertex of the parabola is given as \((0,0)\) and the focus is \((-2,0)\). Since the focus lies on the negative \(x\)-axis, the axis of the parabola is the \(x\)-axis and the parabola opens towards the negative direction of the \(x\)-axis.

For a parabola whose vertex is at the origin and whose axis is the \(x\)-axis, the standard equation is \(y^{2}=4ax\), where the focus is \((a,0)\).

Comparing the given focus \((-2,0)\) with the standard form, we obtain \(a=-2\). Substituting this value of \(a\) in the standard equation, we obtain

\[ \begin{aligned} a&=-2\\ y^{2}&=4ax\\ y^{2}&=4\left(-2\right)x\\ y^{2}&=-8x \end{aligned} \]

Hence, the required equation of the parabola is \(y^{2}=-8x\).

Q11. Vertex (0,0) passing through (2,3) and axis is along x-axis.

Solution

The vertex of the parabola is given as \((0,0)\) and its axis is the \(x\)-axis. Hence, the parabola opens along the \(x\)-axis and its standard equation is \(y^{2}=4ax\).

Since the parabola passes through the point \((2,3)\), the coordinates of this point must satisfy the equation of the parabola. Substituting \(x=2\) and \(y=3\) in \(y^{2}=4ax\), we obtain

\[ \begin{aligned} 3^{2}&=4a\cdot 2\\ 9&=8a\\ a&=\dfrac{9}{8} \end{aligned} \]

Substituting the value of \(a=\dfrac{9}{8}\) in the standard equation \(y^{2}=4ax\), the equation of the required parabola is

\[ \begin{aligned} y^{2}&=4\cdot a\cdot x\\ y^{2}&=4\cdot \dfrac{9}{8}x\\ y^{2}&=\dfrac{9}{2}x\\ 2y^{2}&=9x \end{aligned} \]

Thus, the equation of the parabola is \(2y^{2}=9x\).

Q12. Vertex (0,0), passing through (5,2) and symmetric with respect to y-axis.

Solution

The vertex of the parabola is \((0,0)\) and it is symmetric with respect to the \(y\)-axis. Hence, the axis of the parabola is the \(y\)-axis and its standard equation is \(x^{2}=4ay\).

Since the parabola passes through the point \((5,2)\), the coordinates of this point must satisfy its equation. Substituting \(x=5\) and \(y=2\) in \(x^{2}=4ay\), we obtain

\[ \begin{aligned} 5^{2}&=4\cdot a\cdot 2\\ 25&=8a\\ a&=\dfrac{25}{8} \end{aligned} \]

Substituting the value of \(a=\dfrac{25}{8}\) in the standard equation \(x^{2}=4ay\), the equation of the required parabola is

\[ \begin{aligned} x^{2}&=4ay\\ x^{2}&=4\cdot \dfrac{25}{8}y\\ x^{2}&=\dfrac{25}{2}y\\ 2x^{2}&=25y \end{aligned} \]

Thus, the equation of the parabola is \(2x^{2}=25y\).