Conic Sections Exercise 02
Question:
Find the equation of the parabola with focus (6, 0) and directrix x = 6. Also find the length of latusrectum
Answer:
Answer:

The equation of the parabola is given by (y  0)^2 = 4p(x  6), where p is the distance between the focus and directrix.

The length of the latusrectum is 4p.
Question:
Find the coordinates of the focus axis of the parabola, the equation of directrix and the length of the latus rectum for the parabola x^2 =−16y.
Answer:
Step 1: To find the coordinates of the focus axis of the parabola, use the formula (h, k) = (0, 4).
Step 2: To find the equation of the directrix, use the formula y = 4.
Step 3: To find the length of the latus rectum, use the formula l = 4√2.
Question:
Find the coordinates of the focus, axis of the parabola ,the equation of directrix and the length of the latus rectum for x^2=−9y
Answer:

The coordinates of the focus of the parabola is (0, 9/4).

The axis of the parabola is the xaxis.

The equation of the directrix is y = 9/4.

The length of the latus rectum is 9.
Question:
Find the equation of the parabola whose Vertex (0,0); Focus (3,0)
Answer:
Answer: Step 1: Find the equation of the parabola in the form y = ax2 + bx + c.
Step 2: Find the value of “a” using the given information.
The vertex of the parabola is (0,0), which means that the equation of the parabola must have a cvalue of 0. Therefore, the equation of the parabola can be written as y = ax2 + bx.
Step 3: Find the value of “b” using the given information.
The focus of the parabola is (3,0). This means that the equation of the parabola must have a bvalue of 3. Therefore, the equation of the parabola can be written as y = ax2  3x.
Step 4: Find the value of “a” using the given information.
The focus of the parabola is (3,0). This means that the equation of the parabola must have an avalue of 1/4. Therefore, the equation of the parabola can be written as y = (1/4)x2  3x.
Therefore, the equation of the parabola whose vertex is (0,0) and focus is (3,0) is y = (1/4)x2  3x.
Question:
Find the equation of parabola with vertex (0,0) & focus at (0,2).
Answer:
Step 1: Identify the equation of a parabola in standard form. This is y = ax^2 + bx + c.
Step 2: Identify the vertex of the parabola. The vertex is (0,0).
Step 3: Identify the focus of the parabola. The focus is (0,2).
Step 4: Use the vertex and focus to calculate the value of the equation’s constants. The equation of the parabola is y = (1/4)x^2 + 0x + 0.
Question:
Find the equation of the parabola that satisfies the following conditions: Vertex (0,0) passing through (2,3) and axis is along xaxis
Answer:
Answer: Step 1: Identify the vertex of the parabola. Vertex = (0,0)
Step 2: Identify the direction of the axis of the parabola. Axis = xaxis
Step 3: Identify the point that the parabola passes through. Point = (2,3)
Step 4: Substitute the values in the general equation of the parabola: y = ax^2 + bx + c
Step 5: Solve the equation for a, b, and c by substituting the values of the vertex and the point.
Substituting the vertex (0,0): 0 = a(0)^2 + b(0) + c c = 0
Substituting the point (2,3): 3 = a(2)^2 + b(2) + 0 3 = 4a + 2b
Step 6: Solve the simultaneous equations for a and b. 3 = 4a + 2b 4a + 2b = 0
Subtracting 4a from both sides: 2b = 4a b = 2a
Substituting b = 2a in the equation 3 = 4a + 2b: 3 = 4a + 2(2a) 3 = 4a  4a 3 = 0
a = 0.75 b = 1.5
Step 7: The equation of the parabola is: y = 0.75x^2  1.5x
Question:
Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the rectum for y^2=12x
Answer:

The coordinates of the focus are (6, 0).

The axis of the parabola is the xaxis.

The equation of the directrix is x = 2.

The length of the rectum is 12.
Question:
Find the coordinates of the focus, axis of the parabola ,the equation of directrix and the length of the latus reactum for x^2=6y
Answer:

The coordinates of the focus of the parabola is (0,3).

The axis of the parabola is the xaxis.

The equation of the directrix is x = 3.

The length of the latus rectum is 6.
Question:
Find the coordinates of the focus axis of the parabola the equation of directrix and the length of the latus rectum for y^2=−8x
Answer:

The equation of the parabola is y^2=8x.

The coordinates of the focus axis is (0, 4).

The equation of the directrix is x=1/4.

The length of the latus rectum is 8.
Question:
Find the equation of the parabola that satisfies the following conditions: Vertex (0,0) passing through (5,2) and symmetric with respect to y axis
Answer:
Step 1: Since the vertex is (0,0) and the parabola is symmetric with respect to the yaxis, the equation of the parabola should be in the form of y = ax^2.
Step 2: Substitute the given coordinates in the equation and solve for a.
y = a(x^2)
0 = a(0^2)
2 = a(5^2)
2/25 = a
Step 3: Substitute the value of a in the equation and solve for y.
y = (2/25)x^2
Therefore, the equation of the parabola is y = (2/25)x^2.
Question:
Find the coordinates of the focus axis of the parabola the equation of directrix and the length of the latus rectum for y^2=10x
Answer:
Answer: Step 1: Find the equation of the vertex: The equation of the vertex of the parabola is (0, 0).
Step 2: Find the equation of the directrix: The equation of the directrix of the parabola is x = 5.
Step 3: Find the coordinates of the focus axis: The coordinates of the focus axis of the parabola is (5, 0).
Step 4: Find the length of the latus rectum: The length of the latus rectum is 10 units.
Question:
Find the equation of the parabola that satisfies the following conditions : Focus(0,3) and directrix y=3.
Answer:
Answer: Step 1: The equation of a parabola with focus (0,3) and directrix y=3 is given by the formula: y^2 = 4px + 8p
Step 2: Substituting the given values in the formula, we get: y^2 = 4p(0) + 8p
Step 3: Simplifying the equation, we get: y^2 = 8p
Step 4: Thus, the equation of the parabola satisfying the given conditions is y^2 = 8p.