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 y² = 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 x² = 4ay
Arch is 10 m high and 5 m wide at base.

∴ Equation of parabola becomes x² = (frac{5}{8})y.
Let width of arch 2 m from vertex is 2b, then, point (b, 2) lies on parabola.
b² = (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 x² = 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 x² = (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 x² + 4y² = 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 + 4y² = 16.
⇒ 4y² = 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 OA² + OB² = AB², 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 x² = 12y to the ends of its latus rectum.
Solution:
Parabola is x² = 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 y² = 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 y² = 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.