Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 11 Conic Sections Miscellaneous Exercise Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 11 Conic Sections Miscellaneous Exercise

Question 1.

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

Solution:

Taking vertex of the parabolic reflector at origin, x-axis along the axis of parabola, the equation of parabola is y^{2} = 4ax.

Given depth 5 cm, diameter 20 cm.

∴ (5, 10) point lies on parabola.

(10)^{2} = 4a(5)

∴ a = 5.

∴ Focus is (a, 0), i.e., (5, 0), which is the mid-point of the given diameter.

Question 2.

An arch is in the form olf 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?

Solution:

∵ Axis of the parabola is vertical.

∴Its equation is x^{2} = 4ay

Arch is 10 m high and 5 m wide at base.

∴ Equation of parabola becomes x^{2} = \(\frac{5}{8}\)y.

Let width of arch 2 m from vertex is 2b, then, point (b, 2) lies on parabola.

b^{2} = \(\frac{5}{8}\).2 = \(\frac{5}{4}\)

∴ b = \(\frac{\sqrt{5}}{2}\).

∴ Width of arch is 2b = 2.\(\frac{\sqrt{5}}{2}\)m = \(\sqrt{5}\)m = 2.23 m (approx.).

Question 3.

The cable of a uniformly loaded suspension bridge hangs 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.

Solution:

The cable is in the form of a parabola x^{2} = 4ay.

Focus is at the middle of the cable and, shortest and longest vertical supports are 6 m and 30 m, and roadways is 100 m long.

∴ Point (50, 24) lies on the parabola.

∴ (50)^{2} = 4a(2a) or 4a = \(\frac{625}{6}\).

∴ Equation of parabola is x^{2} = \(\frac{625}{6}\)y.

Let the support at 18 m from middle be B m. Then, (18, B – 6) lies on the parabola

= 9.11 (approx.)

∴ Length of support is 9.11 m (approx.)

Question 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.

Solution:

Since the arch is 8 m wide and 2 m high at centre in the form of a semi-ellipse, so

arch is a part of ellipse whose semi-major and semi-minor axes are 4 m and 2 m respectively.

i.e., a = 4, b = 2.

∴ Its equation is \(\frac{x^{2}}{16}\) + \(\frac{y^{2}}{4}\) = 1

or x^{2} + 4y^{2} = 16.

A point 1.5 m from one end of arch is (4 – 1.5) m = 2.5 m from origin.

Height of the arch at this point is given by value of it obtained by putting x = 2.5.

∴ (2.5)^{2} + 4y^{2} = 16.

⇒ 4y^{2} = 16 – 6.25 = 9.75

⇒ y = \(\frac{\sqrt{9.75}}{2}\) = \(\frac{3.12}{2}\) = 1.56 (approx.)

∴ Height of the arch is 1.56 m (approx.)

Question 5.

A rod of length 12 cm moves with its ends always touching the co-ordinate 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 x-axis.

Solution:

Let AB = 12 cm is the rod and P(x, y) is point on the rod such that

PA = 3 cm.

∴ PB = 9 cm.

Draw PQ ⊥ OB and PR ⊥ OA.

Let BQ = b and AR = a.

Then, from similar ∆s BQP and PRA, we have:

\(\frac{b}{9}\) = \(\frac{y}{3}\) and \(\frac{a}{3}\) = \(\frac{x}{9}\).

⇒ b = 3y and a = \(\frac{1}{3}\)x.

∴ OA = x + a = x + \(\frac{1}{3}\)x = \(\frac{4}{3}\)x

and OB = y + b = y + 3y = 4y.

Using OA^{2} + OB^{2} = AB^{2}, we have:

which is locus of P.

Question 6.

Find the area of the triangle formed by the lines joining the vertex of the parabola x^{2} = 12y to the ends of its latus rectum.

Solution:

Parabola is x^{2} = 12y.

∴ Length of latus rectum AB is 4a = 12.

Also ⊥ distance p = ON of vertex O from latus rectum AB is given by

(6)^{2} = 12p or p = 3.

∴ Area of ∆ formed by joining vertex

O the ends A and B of latus rectum is the area of ∆ OAB

= \(\frac{1}{2}\) × 12 × 3 = 18 sq. units.

Question 7.

A man running a race-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 post is 8 m.

Find the equation of the path traced by the man.

Solution:

The path traced by the man according to the given condition will be an ellipse,

whose foci S and S’ will be the flag posts and the sum of distances of man P from S and S’ is equal to major axis.

If \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 is eqn. of the ellipse,

then, PS + PS’ = 2a = 10.

Now, since a = 5 m so

SS’ = 2ae = 8.

Question 8.

An equilateral triangle is inscribed in the parabola y^{2} = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Solution:

Equation of parabola is y^{2} = 4ax.

Let p is the side of the equilateral AOAB whose one vertex is at the vertex of parabola.

Then, by symmetry AB is ⊥ to the axis ON of parabola

∴ Side of the traiangle is 8\(\sqrt{3}\)a.