Test: Conic Sections: General Equations of Conics (19 July) - JEE MCQ

Formula Used:
(a + b)² = a² + 2ab + b²
Calculation:
x² + 4x - 2y + 5 = 0 
x² + 4x + 4  - 2y + 1 = 0 
Since, (a + b)² = a² + 2ab + b²
(x + 2)² - (2y - 1) = 0
(x + 2)² = (2y - 1) 
This will represent the parabola which is not given in any of the options. 
∴ The correct answer will be None of the above.

Formula Used:
(a + b)² = a² + 2ab + b²
Calculation:
x² + 4x - 2y + 5 = 0 
x² + 4x + 4  - 2y + 1 = 0 
Since, (a + b)² = a² + 2ab + b²
(x + 2)² - (2y - 1) = 0
(x + 2)² = (2y - 1) 
This will represent the parabola which is not given in any of the options. 
∴ The correct answer will be Neither of the above.

Given:

Squaring both sides,

Squaring both sides,
⇒ 256 + 256x + 64x² = 64[x² + 4x + 4 + y²]
⇒ 4 + 4x + x² = [x² + 4x + 4 + y²]
⇒ y² = 0 or y = 0, y = 0
which is the equation of pair of straight lines
∴ The correct answer is option (4).

Given:
A(3, 4, 5) and B(-1, 3, -7) are two points.
Formula Used:
Distance formula = 
Calculation:
Let the coordinates of point P be (x, y, z)
We need to find the equation of the set of point P(x, y, z) such that PA² + PB² = 2n²    ------ equation (1)
Calculating (PA)²
P(x, y, z) and A(3, 4, 5)
Here, x₁ = x, y₁ = y, z₁ = z, x₂ = 3, y₂ = 4, z₂ = 5

Squaring on both sides,

⇒ (PA)² = (3 - x)² + (4 - y)² + (5 - z)²
⇒ (PA)² = 9 + x² - 6x + 16 + y² - 8y + 25 + z² - 10z 
⇒ (PA)² = x² + y² + z² - 6x - 8y - 10z + 50
Calculating (PB)²
P(x, y, z) and B(-1, 3, -7)
Here, x₁ = x, y₁ = y, z₁ = z, x₂ = -1, y₂ = 3, z₂ = -7

Squaring on both sides,

⇒ (PB)² = (-1 - x)² + (3 - y)² + (- 7 - z)²
⇒ (PB)² = 1 + x² + 2x + 9 + y² - 6y + 49 + z² + 14z
⇒ (PB)² = x² + y² + z² + 2x - 6y + 14z + 59
Putting the value of (PA)² and (PB)² in equation (1),
⇒ (PA)² + (PB)² = 2n²
⇒ (x² + y² + z² - 6x - 8y - 10z + 50) + (x² + y² + z² + 2x - 6y + 14z + 59) = 2n²  
⇒ 2x² + 2y² + 2z² - 4x - 14y + 4z + 50 + 59 = 2n²
⇒ 2x² + 2y² + 2z² - 4x - 14y + 4z  = 2n² - 109
∴ Equation of set of points P is 2x² + 2y² + 2z² - 4x - 14y + 4z  = 2n² - 109

Concept:
Hyperbola:
1. The general equation of a hyperbola is  and its conjugate hyperbola is  
2.  The eccentricity e of the hyperbola is  is given by e =  (more than 1).
Calculation:
Let the equation of the hyperbola is

The eccentricity of the hyperbola is 

The equation of its conjugate hyperbola will be 

Adding equation (1)) & (2)

CONCEPT:
The general equation of a non-degenerate conic section is: Ax² + Bxy + Cy² + Dx + Ey + F = 0 where A, B and C are all not zero
The above given equation represents a non-degenerate conics whose nature is given below in the table:

CALCULATION:
Given: 4x² - 9y² + 36x + 36y - 125 = 0
By comparing the given equation with Ax² + Bxy + Cy² + Dx + Ey + F = 0, we get
⇒ A = 4, B = 0, C = - 9, D = 36, E = 36 and F = - 125
Here, we can see that, B = 0, A ≠ C and sign of A and C are opposite
As we know that, if B = 0, A ≠ C and sign of A and C are opposite then the non-degenerate equation represents a hyperbola
Hence, option D is the correct answer.

Concept:
The general equation of a non-degenerate conic section is: ax² + 2hxy + by² + 2gx + 2fy + c = 0 where a, h and b are all not zero
The above-given equation represents a non-degenerate conics whose nature is given below in the table:

To convert from Polar Coordinates (r, θ) to Cartesian Coordinates (x, y)

• x = r × cos (θ)
• y = r × sin (θ)

To convert from Cartesian Coordinates (x, y) to Polar Coordinates (r, θ)

Calculation:

⇒ 3 = r + 2 r cos θ
⇒ 3 = r + 2x            [∵ r cos θ = x]
⇒ 3 - 2x = r
Squaring both sides, we get
⇒ (3 - 2x)² = r²
⇒ 9 + 4x² - 12x = x² + y²     [∵ r² = x2 + y²]
⇒ 3x² - y² - 12x + 9 = 0
By comparing the given equation with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get
⇒ a = 3, h = 0, b = -1
Here, we can see that, h = 0, a ≠  b and sign of a and b are opposite.
Hence,  equation represents a hyperbola.
Alternate Method
Polar form of conic section: 1/r = 1 + ecosθ
Given equation is 
Here e = 6, which is greater than zero,
Hence, the given equation is a hyperbola.
Additional Information

• Eccentricity of ellipse 0 < e < 1
• Eccentricity of parabola e = 1
• Eccentricity of hyperbola e > 1
• Eccentricity of circle e = 0

CONCEPT:
The general equation of a non-degenerate conic section is: Ax² + Bxy + Cy² + Dx + Ey + F = 0 where A, B and C are all not zero
The above given equation represents a non-degenerate conics whose nature is given below in the table:

CALCULATION:
Given: 9y² + 16x + 36y - 10 = 0
By comparing the given equation with Ax² + Bxy + Cy² + Dx + Ey + F = 0, we get
⇒ A = 0, B = 0, C = 9, D = 16, E = 36 and F = - 10
Here, we can see that, B = 0 and A = 0
As we know that, if B = 0 and either A = 0 or C = 0 then the non-degenerate equation represents a parabola.
Hence, option A is the correct answer.

Concept:
The general equation of a cone is given by
ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0
Where a, b, c, d, e, f are real numbers and a ≠ 0, b ≠ 0, c ≠ 0.
The point on three coordinate axes x, y, and z are
(x, 0, 0), (0, y, 0), and (0, 0, z).
Explanation:
The general equation of a cone with a vertex at the origin is given by
ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0    ----(1)
According to the question, the cone passes coordinate axes, it must satisfy 
(x, 0, 0), (0, y, 0), and (0, 0, z).
From here we will get that
a = 0, b = 0 and c = 0
Therefore, from equation (1)
2fyz + 2gzx + 2hxy = 0
⇒ fyz + gzx + hxy = 0      -----(2)
Given that the lines 
are the generators of the cone, thus d.r.s of the lines satisfy the equation.
f(−2)(3) + g(3)(1) + h(1)(−2) = 0
⇒ −6f + 3g − 2h = 0       ------(3)
f(−1)(−1) + g(1)(3) + h(3)( −1) = 0
⇒ f − 3g − 3h = 0         -----(4)

⇒ f = −3k, g = −16k, h = -15k
From equation (3)
⇒ −3kyz − 16kzx - 15kxy = 0
∴  3yz + 16zx + 15xy = 0

Concept:
The general equation of a non-degenerate conic section is: ax² + 2hxy + by² + 2gx + 2fy + c = 0 where a, h and b are all not zero
The above-given equation represents a non-degenerate conics whose nature is given below in the table:

Calculation:
x = a cosα − b sinα        ....(i) 
y = a sinα − b cosα        ....(ii)
Squaring both sides of (i) and (ii) and adding both the equation we get  x² + y² = a² + b².
By comparing the given equation with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get
⇒ a = 1, h = 0, b = 1
Here, we can see that h = 0, a = b
Which represents the locus of the circle.