# NCERT Solutions class-11 Maths Miscellaneous

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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 F1 are and F2 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

OA2 = OC2 + AC2

Therefore, the side of triangle is