Conic Sections Miscellaneous Exercise
Question:
A rod of length 12 cm moves with it ends always touching the coordinate axes. Dertermine the equation of the locus of a point P on the rod which is 3 cm from the end in contact with the xaxis
Answer:
Answer:
 Let the point P be (x, y).
 The length of the rod is 12 cm, so the coordinates of the other end of the rod, in contact with the xaxis, will be (12, 0).
 The distance between P and the other end of the rod is 3 cm, so we can use the distance formula to calculate the coordinates of P: d = √((x  12)2 + y2) = 3
 Solving for x and y, we get: x = 9 and y = √(9)
 Therefore, the equation of the locus of P is: x = 9 and y = √(9)
Question:
A man running a race course notes that the sum of the distance 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.
Answer:
Let the flag posts be A and B, and the distance from the man to A and B be x and y respectively.
Given: Distance from man to A + Distance from man to B = 10 m Distance between A and B = 8 m
Equation: x + y = 10 x  y = 8
Solving the equations: x + y = 10 x = 10  y
Substituting x = 10  y in the second equation, (10  y)  y = 8 10  2y = 8 2y = 2 y = 1
Substituting y = 1 in the first equation, x + 1 = 10 x = 9
Therefore, the equation of the posts traced by the man is x = 9 and y = 1.
Question:
An arch is in the form of a semiellipse. 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.
Answer:
Step 1: Draw a diagram of the arch.
Step 2: Label the diagram. Label the width as 8 m, the height at the centre as 2 m, and the distance from one end as 1.5 m.
Step 3: Use the formula for the equation of an ellipse to calculate the height of the arch at 1.5 m from one end. The formula is:
Height = 2 * sqrt(1  (x/a)^2) * b
where a is the width of the arch (8 m) and b is the height at the centre (2 m).
Step 4: Substitute the values into the equation:
Height = 2 * sqrt(1  (1.5/8)^2) * 2
Step 5: Simplify the equation:
Height = 2 * sqrt(0.6875) * 2
Step 6: Calculate the height of the arch:
Height = 2 * 0.83 * 2
Step 7: The height of the arch at 1.5 m from one end is 3.32 m.
Question:
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 form the vertex of the parabola?
Answer:
Step 1: Identify the equation of the parabola: The equation of the parabola is y = ax^2 + bx + c, where a, b, and c are constants.
Step 2: Substitute the given values into the equation: The arch is 10 m high and 5 m wide at the base, so we can substitute these values into the equation: 10 = a(5^2) + b(5) + c
Step 3: Solve for the constants a, b, and c: Solving for a, b, and c, we get: a = 0.2, b = 0, c = 10
Step 4: Substitute the value of x = 2 into the equation: Substituting x = 2 into the equation, we get: 10 = 0.2(2^2) + 0(2)  10
Step 5: Solve for y: Solving for y, we get: y = 4 m
Question:
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.
Answer:

Find the equation of the parabola that describes the shape of the cable.

Find the equation of the line that describes the roadway.

Find the equation of the line that describes the vertical wire attached to the cable.

Substitute 18 m for the xvalue in the equation of the line.

Solve for the corresponding yvalue.

Use the equation of the parabola to substitute the yvalue into it and solve for the corresponding xvalue.

Use the equation of the line to substitute the xvalue into it and solve for the corresponding yvalue.

The yvalue is the length of the supporting wire attached to the roadway 18 m from the middle.
Question:
If a parabola reflector with its axis as x−axis is 20 cm in diameter and 5 cm deep. Find the focus.
Answer:
Step 1: A parabola reflector is a parabolic mirror.
Step 2: The formula for the focus of a parabola is F = 4p * d^2/16 * h, where d is the diameter, h is the depth and p is the constant pi.
Step 3: Substitute the given values in the formula to calculate the focus, F = 4p * (20 cm)^2/16 * (5 cm) = 50 cm.
Therefore, the focus of the parabola reflector is 50 cm.
Question:
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.
Answer:
Given: Parabola equation: x2 = 12y
Step 1: Find the coordinates of the vertex of the parabola.
The vertex of the parabola can be found by taking the derivative of the equation and setting it equal to zero.
Derivative of x2 = 12y 2x = 12
Setting it equal to zero: 2x = 0 x = 0
Therefore, the vertex of the parabola is (0, 0).
Step 2: Find the length of the latus rectum.
The latus rectum is the line that passes through the vertex and is perpendicular to the directrix of the parabola. The directrix of the parabola is x = 6.
Therefore, the length of the latus rectum is 6 units.
Step 3: Find the coordinates of the ends of the latus rectum.
The latus rectum is 6 units in length, so the coordinates of the ends of the latus rectum can be found by adding and subtracting 3 units from the xcoordinate of the vertex.
The coordinates of the ends of the latus rectum are (3, 0) and (3, 0).
Step 4: Find the area of the triangle formed by the lines joining the vertex of the parabola to the ends of its latus rectum.
The area of the triangle can be found using the formula for the area of a triangle:
Area = 1/2 × base × height
The base of the triangle is 6 units (the length of the latus rectum) and the height is 0 (since the triangle is formed by lines joining the vertex of the parabola to the ends of its latus rectum).
Therefore, the area of the triangle is 0.