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Applied Maths Chapter 18 Solutions
Circle And Parabola
EXERCISE- 18.2
Q.1 Find the length of the latus-rectum of each of the following parabolas:
Ans.
(i) y^2 = 8x
y^2 = 4ax
here, 4a = 8
Length of lactus rectum = 8
(ii) y^2= -12x
y^2 = -4ax
here, -4a = -12
Length of lactus rectum = 12
(iii) x^2 = 16y
x^2 = 4ay
here, 4a = 16
Length of lactus rectum = 16
(iv) x^2=-10y.
x^2 = 4ay
here, -4a = -10
Length of latus rectum = 10
Q.2 Find the coordinates of the focus of each of the following parabolas:
Ans.
(i) y^2 = 12x
y^2 = 4ax
here, 4a = 12
a = 3
Therefore, form = (3, 0)
(ii) y^2= -8x
y^2 = -4ax
here, -4a = -8
a = 2
Therefore, form = (-2, 0)
(iii) x^2 = 16y
x^2 = 4ay
here, 4a = 16
a = 4
Therefore, form = (0, 4)
(iv) x^2= -9y
y^2 = -4ax
here, -4a = -9
a = 9/4
Therefore, form = (0, -9/4)
Q.3 In each of the following parabolas, find the coordinates of focus, the axis of the parabola, the equation of the directrix, and the length of the latus-rectum:
Ans.
(i) y^2 = 12x
y^2 = 4ax
4a = 12
a = 3
Focus = (3, 0)
Axis = x-axis or y = 0
Equation of directrix,
x = -3, or x + 3 = 0
Length of lactus rectum = 4a
= 12
(ii) y^2= -8x
y^2 = -4ax
-4a = -8
a = 2
Focus = (-2, 0)
Axis = x-axis or y = 0
Equation of directrix,
x = 2, or x – 2 = 0
Length of lactus rectum = 4a
= 8
(iii) x^2 = 6y
x^2 = 4ay
4a = 6
a = 3/2
Focus = (0, 3/2)
Axis = y-axis or x = 0
Equation of directrix,
x = -3/2, or 2y + 3 = 0
Length of lactus rectum = 4a
= 6
(iv) x^2= -8y
x^2 = -4ay
-4a = -8
a = 2
Focus = (0, -2)
Axis = y-axis or x = 0
Equation of directrix,
y = 2, or y – 2 = 0
Length of latus rectum = 4a
= 8
Q.4 Find the coordinates of the endpoints of the latus-rectum of the parabola y^2 = 8x.
Ans. y^2 = 8x
y^2 = 4ax
4a = 8
a = 2
Therefore, Focus (2, 0)
Now, because the parabola passes through (2, k)
k^2 = 8 x 2
k^2 = 16
k = ±4
so, the end point of the latus rectum are (2, 4) (2, -4)
Q.5 Find the equation of the parabola with Also,
(i) focus at (0, 4) and directrix y + 4 = 0
(ii) focus at (0,-3) and directrix y = 3.
find the length of the latus-rectum in each case.
Ans.
(i) focus at (0, 4) and directrix y + 4 = 0
Because focus is (0, 4)
Therefore, parabola axis is y-axis
x^2 = 4ay
x^2 = 4 (4)y
x^2 = 16y
Length of lactus rectum = 4a
= 16
(ii) focus at (0,-3) and directrix y = 3.
Because the focus is (0, -3)
Therefore, the parabola axis is the y-axis
x^2 = 4ay
x^2 = 4 (-3)y
x^2 = -12y
Length of latus rectum = -4a
= 12
Q.6 If the parabola y^2 = 4px passes through the point (3, 2), find the length of the latus-rectum and the coordinates of the focus.
Ans. y^2 = 4px , (3, 2)
(2)^2 = 4p (3)
4 = 12p
p = 4/12 = 1/3
Therefore, equation of parabola
y^2 = 4ax
y^2 = 4 X 1/3 x
3 y^2 = 4x
So, a = 1/3
Length of lactus rectum = 4a
= 4 x 1/3
= 4/3
Coordinate of focus = (1/3, 0)
Q.7 Find the equation of the parabolas with vertices at the origin and satisfying the following conditions:
(i) Focus at (3,0)
(ii) Focus at (0,2)
(iii) Focus at (0,-4)
(iv) Directrix y – 2 = 0
(v) Passing through (2, 3) and axis along the x-axis
(vi) Passing through (5, 2) and symmetric concerning the y-axis
Ans.
(i) Focus at (3,0)
y^2 = 4ax
y^2 = 4 (3)x
y^2 = 12x
(ii) Focus at (0,2)
x^2 = 4ay
x^2 = 4 (2)y
x^2 = 8y
(iii) Focus at (0,-4)
x^2 = 4ay
x^2 = 4 (-4)y
x^2 = -16y
(iv) Directrix y – 2 = 0
y – 2 = 0
y = 2
Therefore, a = -2
Equation of parabola, x^2 = 4ay
x^2= 4(-2)y
x^2 = -8y
(v) Passing through (2, 3) and axis along the x-axis
y^2 = 4ax
3^2 = 4a(2)
9 = 8a
a = 9/8
Equation of parabola, y^2 = 4ax
= 4 x 9/8 x
= 9/2x
2y^2 = 9x
(vi) Passing through (5, 2) and symmetric concerning the y-axis
x^2 = 4ay
5^2 = 4a(2)
25 = 8a
a = 25/8
Equation of parabola, x^2 = 4ay
= 4 x 25/8 y
= 25/2y
2x^2 = 25y
Q.8 Find the equation of a parabola whose focus is the point (2, 3) and directrix is the line x – 4y + 3 = 0.
Ans.
MP = FP …………(i)
MP = |Ax1 + Bx2 + C|/√A^2 +B^2
= |1(x) + (-4)(y) + 3|/√(1)^2 + (-4)^2
= |x – 4y + 3|/√17
FP = √(x-1)^2 + (y-3)^2
from eq.(i)
MP^2 = FP^2
(x – 4y + 3)^2/17 = (x – 2)^2 + (y – 3)^2
x^2 + 16y^2 + 9 – 8xy – 24y + 6x = 17 [x^2 – 4x + 4 + y^2 – 6y + 9]
x^2 + 16y^2 – 8xy – 24y + 6x + 9 = 17x^2 – 68x + 68 + 17y^2 + 102y + 153
16x^2 + y^2 + 8xy – 74x – 78y + 221 – 9 = 0
16x^2 + y^2 + 8xy – 74x – 78y + 212 = 0
FAQ’s related to Applied Maths Chapter 18 on Circle And Parabola:
Q.1 What is the equation of a circle and how is it derived?
Ans. The equation of a circle is (𝑥−ℎ)^2 + (𝑦−𝑘)^2= r^2, where (ℎ, 𝑘)(h, k) is the center of the circle and 𝑟 is the radius. This equation is derived from the distance formula between two points.
Q.2 What are the standard forms of the equation of a parabola?
Ans. The standard forms of the equation of a parabola are:
- Vertical Axis: 𝑦 = 𝑎𝑥^2 + 𝑏𝑥 + c
- Horizontal Axis: 𝑥 = 𝑎𝑦^2 + 𝑏𝑦 + 𝑐
These are few Frequently Asked Questions relating to Applied Maths Chapter 18
In Applied Maths chapter 18, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Applied Maths Chapter 18 promises an enriching learning experience that will set you on the path to success. Applied Maths Chapter 18, we delve deep into advanced mathematical concepts that are crucial for understanding.
