**Q.1: For the equation y ^{2} = 16x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.**

** **

**Sol:**

**Given:**

** y ^{2} = 16x**

As the coefficient of x is positive, the given **parabola has the opening on the right side.**

By comparing this equation with y^{2} = 4ax, we get:

4a = 16 ⇒ **a = 4**

Therefore, **focus coordinates** = (a, 0) **= (4, 0)**

The given equation has y^{2}, the x-axis is the axis of the parabola.

Directrix equation, x = – a = – 4

So, **x + 4 = 0**

**Therefore, Length of the latus rectum = 4a = 4 × 4 = 16**

**Q.2: For the equation x ^{2} = 8y, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.**

** **

**Sol:**

**Given:**

** x ^{2} = 8y**

As the coefficient of y is positive, the given** parabola has the opening upwards.**

By comparing this equation with x^{2} = 4ay, we get

4a = 8 ⇒ **a = 2**

Therefore, **focus coordinates** = (0, a) **= (0, 2)**

The given equation has x^{2}, the y – axis is the axis of the parabola.

Directrix equation, y = – a = – 2

So, **x + 2 = 0**

**Therefore, Length of the latus rectum = 4a = 4 × 2 = 8**

** **

**Q.3: For the equation y ^{2} = – 12x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.**

** **

**Sol:**

**Given:**

** y ^{2} = – 12x**

As the coefficient of x is negative, the given **parabola has the opening on the left side.**

By comparing this equation with y^{2} = – 4ax, we get

4a = 12 ⇒ **a = 3**

Therefore, **focus coordinates** = (- a, 0) **= (- 3, 0)**

The given equation has y^{2}, the x-axis is the axis of the parabola.

** Directrix equation, x = a = 2**

So,** x + 4 = 0**

**Therefore, Length of the latus rectum = 4a = 4 × 3 = 12**

** **

**Q.4: For the equation x ^{2} = – 20y, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.**

** **

**Sol:**

**Given:**

** x ^{2} = – 20y**

As the coefficient of y is negative, the given **parabola has the opening downwards.**

By comparing this equation with x^{2} = – 4ay, we get:

– 4a = – 20 ⇒ **a = 5**

Therefore, **focus coordinates** = (0, a) = **(0, 5)**

The given equation has x^{2}, the y – axis is the axis of the parabola.

** Directrix equation, y = a = 5**

**Therefore, Length of the latus rectum = 4a = 4 × 5 = 20**

** **

**Q.5: For the equation y ^{2} = 24x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.**

** **

**Sol:**

**Given:**

** y ^{2} = 24x**

As the coefficient of x is positive, the given **parabola has the opening on the right side.**

By comparing this equation with y^{2} = 4ax, we get:

4a = 24 ⇒ **a = 6**

Therefore, **focus coordinates**** =** (a, 0) = **(6, 0)**

The given equation has y^{2}, the x-axis is the axis of the parabola.

**Directrix equation, x = – a = – 6**

**So, x + 6 = 0**

**Therefore, Length of the latus rectum = 4a = 4 × 6 = 24**

**Q.6: For the equation x ^{2} = – 7y, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.**

**Sol:**

**Given:**

** x ^{2} = – 7y**

As the coefficient of y is negative, the given **parabola has the opening downwards.**

By comparing this equation with x^{2} = – 4ay, we get:

– 4a = – 7 ⇒ a =

Therefore, **focus coordinates** = (0, a) = (0,

The given equation has x^{2}, **the y – axis is the axis of the parabola**.

** Directrix equation,** y = a =

**Therefore, Length of the latus rectum = 4a = 4 × 74 = 7**

** **

**Q7. Obtain the equation of the parabola which satisfies the given conditions below: **

**(i) Focus coordinates**** = (8, 0) **

**(ii) Directrix (x) = – 8**

** **

**Sol:**

**Given:**

**(i) Focus coordinates**** = (8, 0)**

**(ii) Directrix (x) = – 8**

The axis of the given parabola is the x – axis as the focus is represented on the x – axis.

The required equation can be either y^{2} = 4ax and y^{2} = – 4ax

As we know, **d****irectrix (x) = – 8**

As the coefficient of x is negative, the given **parabola has the opening on the** **left side and the focus (8, 0)**

Therefore, the equation of the parabola = y^{2} = 4ax,

**Here, a = 8**

**Thus, the equation of the parabola = y ^{2} = 32 x.**

** **

**Q.8: Obtain the equation of the parabola which satisfies the given conditions below: **

**(i) Focus coordinates**** = (0, – 5)**

**(ii) Directrix (y) = 5**

** **

**Sol:**

**Given:**

**(i) Focus coordinates**** = (0, – 5)**

**(ii) Directrix (x) = 5**

The axis of the given parabola is the y – axis as the focus is represented on the y – axis.

The required equation can be either x^{2} = 4ay and x^{2} = – 4ay

As we know, **d****irectrix (x) = 5**

As the coefficient of x is negative, the given **parabola has the opening on the left side and the focus (0, -5)**

Therefore, the equation of the parabola = x^{2} = – 4ay,

**Here, a = 5**

**Thus, the equation of the parabola = x ^{2} = – 20y**

**Q.9: Obtain the equation of the parabola which satisfies the given conditions below: **

**(i) Vertex is at origin.**

**(ii) Focus coordinates**** = (4, 0)**

** **

**Sol:**

**Given:**

**(i) Vertex is at origin i.e., (0, 0)**

**(ii) Focus coordinates**** = (4, 0)**

As the vertex of a parabola lies at origin and the focus is represented on the positive side of the x – axis, and hence, the axis of the parabola is x – axis.

**The required equation is ****y ^{2} = 4ax **

As we know, focus is (4, 0)

Therefore, the equation of the parabola = y^{2} = 4ax

**Here, a = 4**

**Thus, the equation of the parabola = y ^{2} = 16 x **

** **

**Q.10: Obtain the equation of the parabola which satisfies the given conditions below: **

**(i) Vertex is at origin.**

**(ii) Focus coordinates**** = (- 3, 0)**

** **

**Sol:**

**Given:**

**(i) Vertex is at origin i.e., (0, 0)**

**(ii) Focus coordinates**** = (- 3, 0)**

As the vertex of a parabola lies at origin and the focus is represented on the negative side of the x – axis, and hence, **the axis of the parabola is x – axis.**

The required equation is **y ^{2} = – 4ax**

Therefore, the equation of the parabola = y^{2} = – 4ax

As we know, focus is (- 3, 0)

**Here, a = 3**

**Thus, the equation of the parabola = y ^{2} = – 12 x**

** **

**Q.11: Obtain the equation of the parabola which satisfies the given conditions below: **

**(i) Vertex is at origin.**

**(ii) Passing through coordinates**** = (3, 5)**

**The axis of the parabola is on x – axis.**

**Sol:**

**Given:**

**(i) Vertex is at origin i.e., (0, 0)**

**(ii) Passing through coordinates**** = (3, 5) **

As the vertex of a parabola lies at origin and the passing through coordinates = (3, 5).

The required equation is either y^{2} = 4ax or y^{2} = – 4ax

**The coordinates (3, 5) is in first quadrant**.

Therefore, the equation of the parabola = y^{2} = 4ax, whereas (3, 5) should satisfy the equation

So,

5^{2} = 4a (3)

25 = 12 a

a =

**The parabola’s equation is: **

y^{2} = 4 (

**3 y ^{2} = 25 x**

** **

**Q.12: Obtain the equation of the parabola which satisfies the given conditions below: **

**(i) Vertex is at origin.**

**(ii) Passing through coordinates**** = (6, 4)**

**The equation is symmetric as considered with y – axis.**

**Sol:**

**Given:**

**(i) Vertex is at origin i.e., (0, 0)**

**(ii) Passing through coordinates**** = (6, 4)**

As the vertex of a parabola lies at origin and the passing through coordinates = (6, 4).

The required equation is either x^{2} = 4ay or x^{2} = – 4ay

**The coordinates (6, 4) is in first quadrant.**

Therefore, the equation of the parabola = x^{2} = 4ay, whereas (6, 4) should satisfy the equation

So,

6^{2} = 4a (4)

36 = 16 a

a =

a =

**The parabola’s equation is:**

x^{2} = 4(

**x**^{2}** = 9 y**