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**Miscellaneous Exercise**

**1. In a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.**

**Ans. **A parabolic reflector with diameter PR = 20 cm and OQ = 5 cm

Vertex of parabola is (0, 0)

Let focus of the parabola be

Now, PR = 20 cm

PQ = 10 cm

Coordinate of the point P are (5, 10)

Since the point lies on the parabola

**2. An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?**

**Ans. **Let AB be the parabolic arch having O as the vertex and OY as the axis.

The parabola is of the form

Now, CD = 5 m

OD = 2.5 m

And BD = 10 m

Therefore, coordinates of point B are (2.5, 10).

Since the point B lies on the parabola

Equation of the parabola is

Let PQ =

NQ =

Coordinate of the point Q are

Since the point Q lies on the parabola

Therefore, width of arch is m = 2.24 m approx.

**3. The cable of a uniformly loaded suspension bridge hange in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.**

**Ans. **Let AOB be the cable of uniformly loaded suspension bridge. Let AL and BM be the longest wires of length 30 m each. Let OC be the shortest wire of length 6 m and LM be the roadway.

Now AL = BM = 30 m, OC = 6 m and LM = 100 m

LC = CM = LM = 50 m

Let O be the vertex and axis of the parabola be axis.

Therefore, the equation of the parabola in standard form is

Coordinates of the point B are (50, 24)

Since point B lies on the parabola

Therefore, equation of parabola is

Let length of the supporting wire PQ at a distance of 18 m be

OR = 18 m and PR = PQ – QR = PQ – OC =

Coordinates of point P are

Now, since the point P lies on parabola

9.11 m approx.

**4. An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.**

**Ans. **Given: Width of elliptical arch (AB)

= = 8 = 4 m

Height of the centre (OB) = = 2 m

The axis of ellipse is axis.

Therefore, the equation of ellipse in standard form is .

Now, AP = 1.5 m

OP = OA – AP = 4 – 1.5 = 2.5 m

Let PQ = , then coordinates of Q are

Since the point Q lies on the ellipse

= 1.56 m approx.

**5. A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the ****axis.**

**Ans. **Let AB be a rod of length 12 cm and P be any point on the rod such that PA = 3 cm and PB = 9 cm.

Let AR = and BQ =

Then

And

Now, OA = OR + AR =

And OB = OQ + BQ =

In right angled triangle AOB,

which is required locus of point P and which represents an ellipse.

**6. Find the area of the triangle formed by the lines the vertex of the parabola **** to the ends of its latus rectum.**

**Ans. **Given: Equation of parabola which is in the form of

Focus of the parabola is (0, 3).

Let AB be the latus rectum of the parabola, then

The coordinates of A are

and coordinates of B are (6, 3).

Area of

=

=

= = 18 sq. units

**7. A man running a race leave no space course notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.**

**Ans. **Let be two points where the flag parts are fixed on the ground. The origin O is the mid-point of.

= = 4 m

Coordinates of F_{1} are and F_{2} are (4, 0).

Let P be any point on the track.

Squaring both sides, we have,

Squaring both sides again, we have,

which is the required equation of locus of point P.

**8. An equilateral triangle is inscribed n the parabola **** where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.**

**Ans. **Given: Equation of the parabola .

Let be the side of an equilateral whose one vertex

is the vertex of parabola and let OC =

Now, AB =

AC = BC = =

Coordinates of point A are x’ x

Since, point A lies on the parabola

In right angled triangle

OA^{2} = OC^{2} + AC^{2}

Therefore, the side of triangle is