Unit 3 Emotional Intelligence and Spiritual Intelligence.pdf
Curve Tracing cartesian form.pptx
1. CURVE TRACING
Procedure of tracing of curve in Cartesianform:
I. Symmetry:
a) Symmetric about x-axis:If given equation of curvehaving even
powers of y then the curveis symmetric about x-axis. Or by changing
y to –y if the equation remains unchanged then the curveis
symmetric about x-axis.
Eg. 𝑦2
= 4𝑎𝑥
2. a) Symmetric about y-axis:If given equation of curve having even
powers of x then the curve is symmetric about y-axis. Or by changing
x to –x if the equation remains unchanged then the curveis
symmetric about y-axis.
Eg. 𝑥2
= 4𝑎𝑦
3. (c ) Symmetric about both the axes: if the equation of curveis having
even powers of x as well as even powers of y then the curveis
symmetric about both the axes. Or by changing x to –x and y to –y if the
equation of curveremains unchanged then it is symmetric about both
the axes.
Eg. Equation of circle 𝑥2
+ 𝑦2
= 𝑎2
4. (d) Symmetric about the line y=x: by changing x to y and y to x, if the
equation of curveremains unchanged then the curveis symmetric about
the line y=x
Eg. Equation of circle 𝑥2
+ 𝑦2
= 𝑎2
5. (II) About the origin:
We have to find the curve is passing through the origin or not.
(a) If the equation of curve is free from constant term then the curve is passing
through the origin
Eg. 𝑥2 = 4𝑎𝑦
(b) If the equation of curve is having independent constant term then the curve is not
passing through the origin.
Eg. Equation of circle 𝑥2 + 𝑦2 = 𝑎2
(c) If the curve is passing through the origin then we have to find tangent at the origin
by equating lowest degree term to zero.
8. (III) Points of Intersection:
(a) Points of intersection of the curve with x-axis and y-axis:
By putting y=0 in the given equation of curve we get points of
intersection with x-axis
Similarly by putting x=0 in the given equation of curve we get points of
intersection with y-axis.
(b)If the given equation of curve is symmetric about the line y=x then points of
intersection are calculated by putting y=x and x=y in the given equation of
curve.
9. IV) Asymptote:
Asymptote means the curves which meets at infinite distance.
(a) Equating coefficient of highest powers of x to zero, we get asymptote parallel to x-
axis.
(b) Equating coefficient of highest powers of y to zero, we get asymptote parallel to y-
axis.
Eg 𝑦2 2𝑎 − 𝑥 = 𝑥3
Asymptote parallel to x-axis 1 = 0 𝑖𝑠 𝑛𝑜𝑡 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒
Asymptote parallel to y-axis 2𝑎𝑦2 − 𝑥𝑦2 = 0
2𝑎 − 𝑥 = 0 ⟹ 𝒙 = 𝟐𝒂
10.
11. Trace the curve 𝑥3 + 𝑥𝑦2 − 4𝑦2 = 0
Solution: The curve is symmetric about the x-axis. Because it is having even powers of y.
The curve is passing through the origin because it is not having independent constant
term.
Tangent at the origin:
Equating lowest degree term to zero.
−4𝑦2 = 0
𝑦2 = 0
𝑦 = 0 That is x-axis
Therefore x-axis is tangent at the origin.
16. Trace the curve 𝑦 𝑥2 + 4𝑎4 = 8𝑎3
Solution:
The given equation of curve is 𝑦 𝑥2 + 4𝑎4 = 8𝑎3
The curve is symmetric about y-axis
The curve is not passing through the origin
Because it is having independent constant term.
17. Points of intersection:
𝑦 𝑥2 + 4𝑎4 = 8𝑎3
𝑦𝑥2
+ 4𝑦𝑎4
= 8𝑎3
𝑦𝑥2
+ 4𝑦𝑎4
− 8𝑎3
= 0
Put 𝒙 = 𝟎 𝒕𝒉𝒆𝒏 4𝑦𝑎4 − 8𝑎3 = 0
4𝑎3 𝑎𝑦 − 2 = 0
𝑎𝑦 − 2 = 0
𝑦 =
2
𝑎
The point is (0, 2/a)
Put x=0 then −8𝑎3
= 0 not possible
18. Table
𝑦 =
8𝑎3
𝑥 2 + 4𝑎4
𝐿𝑒𝑡 𝑎 = 1
𝑦 =
8
𝑥 2 + 4
x 0 1 2 3 4 -1 -2 -3 -4
y 2 1.6 1 0.61 0.4 1.6 1 0.61 0.4
Asymptote:
𝑦𝑥 2
+ 4𝑦𝑎4
− 8𝑎3
= 0
Asymptote parallel to x-axis 𝑦 = 0
X-axis is the asymptote parallel to x-axis
Asymptote parallel to y-axis
𝑥 2
+ 4𝑎4
= 0
𝑥 2
+ (2𝑎)2
= 0
𝑥 2
= −(2𝑎)2
𝑥 = ± −(2𝑎)2 Imaginary
Therefore asymptote parallel to y-axis is not possible
21. 1. Trace the curve 𝑦2 𝑎2 + 𝑥2 = 𝑥2 𝑎2 − 𝑥2
a) The curve is symmetric about both the axis.
Because it is having both even powers of x and y
b) The curve is passing through the origin
Because it is not having separate constant term.
𝑦2 𝑎2 + 𝑥2 = 𝑥2 𝑎2 − 𝑥2
𝑦2𝑎2 + 𝑦2𝑥2 = 𝑥2𝑎2 − 𝑥4
𝑦2𝑎2 + 𝑦2𝑥2 − 𝑥2𝑎2 + 𝑥4 = 0
𝑦2𝑎2 − 𝑥2𝑎2 = 0
𝑎2 𝑦2 − 𝑥2 = 0
𝑦2 − 𝑥2 = 0
𝑦 = ±𝑥
Tangent at origin is 𝑦 = ±𝑥
22. (c)Points of intersections
Put 𝑥 = 0 𝑦2𝑎2 = 0 ⟹ 𝑦 = 0
The point is (0,0)
Put 𝑦 = 0 𝑥2 𝑎2 − 𝑥2 = 0
𝑥2 = 0 𝑜𝑟 𝑎2 − 𝑥2 = 0
𝑥 = 0 𝑜𝑟 𝑥 = ±𝑎
The points are (0, 0), (a, 0) & (-a, 0)
Therefore the points (0, 0), (a, 0) & (-a, 0)
(d) Asymptotes
𝑦2𝑎2 + 𝑦2𝑥2 = 𝑥2𝑎2 − 𝑥4
𝑦2
𝑎2
+ 𝑦2
𝑥2
− 𝑥2
𝑎2
+ 𝑥4
= 0
Asymptote parallel to x-axis (1=0) is not possible
Asymptote parallel to y-axis is not possible because
𝑎2
+ 𝑥2
= 0
𝑥2 = −𝑎2
𝑥 = ± −𝑎2 Imaginary
26. 1. Trace the curve𝑥
2
3 + 𝑦
2
3 = 𝑎
2
3
Solution:
a) The curve is symmetric about both the axis. And also it is symmetric
about the line y=x
Because (𝑥)2
1
3 + (𝑦)2
1
3 = (𝑎)2
1
3
b) The curve is not passing through the origin.
Because it is having separateconstant term.
c) Put 𝑥 = 0 (𝑦)2
1
3 = (𝑎)2
1
3 ⇒ 𝑦2
= 𝑎2
⇒ 𝑦 = ±𝑎
The points are (0,a) & (0,-a)
Put 𝑦 = 0 (𝑥)2
1
3 = (𝑎)2
1
3 ⇒ 𝑥2
= 𝑎2
⇒ 𝑥 = ±𝑎
The points are (a,0) & (-a,0)
Therefor The points are (a,0),(-a,0),(0,a) & (0,-a)
d) The asymptote parallel to x & y axis are not possible
Because asymptoteparallel to x-axis 1 = 0 𝑖𝑠 𝑛𝑜𝑡 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒
Asymptoteparallel to y-axis 1 = 0 𝑖𝑠 𝑛𝑜𝑡 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒
27.
28. 2. Trace the curve 𝑦2
2𝑎 − 𝑥 = 𝑥3
with full justification.
a) The curve is symmetric about x-axis
Because it is having even powers of y
b) The curve is passing through the origin
Because it is not having separate constant term.
2𝑎𝑦2
− 𝑥𝑦2
= 𝑥3
2𝑎𝑦2
= 0
𝑦2
= 0
𝑦 = 0 it means x-axis
Tangent at origin: y=0 that is x-axis
29. a) Points of intersection:
𝑝𝑢𝑡 𝑥 = 0 𝑡ℎ𝑒𝑛 𝑦 = 0
𝑝𝑢𝑡 𝑦 = 0 𝑡ℎ𝑒𝑛 𝑥 = 0
The point is (0,0)
Table:
𝑦2
=
𝑥3
2𝑎 − 𝑥
𝑦 = ±
𝑥3
2𝑎 − 𝑥
𝑦 = ±
𝑥3
6 − 𝑥
𝑃𝑢𝑡 𝑎 = 3
x -1 0 1 2 3 4 5 6 7
y Imaginary 0 ± 0.45 ±2 ±3 ±5.65 ±11.18 ∞ Imaginary
For𝒙 ≥ 𝟐𝒂, in this region the curve does not exists and for negative
values of x the curve does not exits
30. d) Asymptote parallel to y-axis is x=2a.
Asymptote parallel to x-axis 1 = 0 𝑖𝑠 𝑛𝑜𝑡 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒
Asymptote parallel to x-axis 2𝑎𝑦2 − 𝑥𝑦2 = 0
2𝑎 − 𝑥 = 0 ⟹ 𝒙 = 𝟐𝒂 (𝒙 = 𝟔
34. 1. Tracing the curve 𝑎𝑦2
= 𝑥3
a) The curve is symmetric about x-axis
Because it is having even power of y
(b) The curve is passing through the origin.
Because it is not having independent constant term
𝑎𝑦2
= 0
𝑦 = 0 (The x-axis)
Tangent at origin: y=0 that is x-axis
35. (c ) Points of intersection:
𝑝𝑢𝑡 𝑥 = 0 𝑡ℎ𝑒𝑛 𝑦 = 0
𝑝𝑢𝑡 𝑦 = 0 𝑡ℎ𝑒𝑛 𝑥 = 0
The point is (0,0)
𝑦2
=
𝑥3
𝑎
𝑦 = ±
𝑥3
𝑎
Take a=5
𝑦 = ± 𝑥3
Table:
x -1 0 1 2 3 4 5
y Imaginary 0 ± 1 ±2.8 ±5.19 ±8 ±11.18
Note:
For negative values of x the curve does not exists
(d) Asymptotes parallel to x and y axis are not possible
38. 5. Trace the curve 𝑦2
𝑎 + 𝑥 = 𝑥2
3𝑎 − 𝑥
Solution: The given equation of curve is 𝑦2
𝑎 + 𝑥 = 𝑥2
3𝑎 − 𝑥
𝑎𝑦2
+𝑥𝑦2
= 3𝑎𝑥2
− 𝑥3
𝑎𝑦2 +𝑥𝑦2 −3𝑎𝑥2 + 𝑥3 = 0 −− − 1
a) The curve is symmetric about x-axis.
Because it is having even powers of x
b) The curve is passing through the origin.
Because it is not having independent constant term.
𝑎𝑦2
− 3𝑎𝑥2
= 0
𝑎 𝑦2 − 3𝑥2 = 0
𝑦2 − 3𝑥2 = 0
𝑦2
= 3𝑥2
𝑦 = ± 3𝑥
Tangent at origin: 𝒚 = ± 𝟑𝒙
39. c) Points of Intersections are (0,0) & (3a,0)
𝑎𝑦2
+𝑥𝑦2
−3𝑎𝑥2
+ 𝑥3
= 0 −− − 1
Put 𝑥 = 0 𝑡ℎ𝑒𝑛 𝑦 = 0
The point is (0, 0)
Put 𝑦 = 0 𝑡ℎ𝑒𝑛
−3𝑎𝑥2 + 𝑥3 = 0
−𝑥2 3𝑎 − 𝑥 = 0
−𝑥2
= 0 𝑜𝑟 3𝑎 − 𝑥 = 0
𝑥 = 0, 𝑥 = 3𝑎
The Points are (0,0) & (3a,0)
Therefore Points of Intersections are (0, 0) & (3a,0)
d) Asymptote parallel to x-axis 1=0 is not possible
Asymptote parallel to y-axis is
𝑎 + 𝑥 = 0
𝒙 = −𝒂
Therefore Asymptote parallel to y-axis is 𝒙 = −𝒂
