General Equation of Conics MCQ Quiz - Objective Question with Answer for General Equation of Conics - Download Free PDF

Condition for Three Mutually Perpendicular Tangent Planes to a Cone:

• The general second-degree equation of a cone is: ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0.
• This form represents a cone with its vertex at the origin.
• Mutually perpendicular planes are three planes intersecting at right angles to each other.
• For a cone to have three mutually perpendicular tangent planes, a special condition must be satisfied.
• Condition: bc + ca + ab = f² + g² + h²
• This condition ensures the cone admits three planes whose normals are mutually perpendicular.
• Tangent Plane: A plane that touches the cone surface at a single point without intersecting it nearby.

Calculation:

Given,

Equation of cone: ax² + by² + cz² + 2ux + 2vy + 2wz + d = 0

Compare this with standard cone: ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0

⇒ To have mutually perpendicular tangent planes, we apply the condition:

⇒ bc + ca + ab = f² + g² + h²

⇒ This relation ensures existence of three mutually perpendicular tangent planes.

∴ Required condition is bc + ca + ab = f² + g² + h²

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:

Equation \(\sqrt {{{(x - 2)}^2} + {y^2}} + \sqrt {{{(x + 2)}^2} + {y^2}} = 4\)

⇒ \(\sqrt {{{(x - 2)}^2} + {y^2}} = 4- \sqrt {{{(x + 2)}^2} + {y^2}} \)

Squaring both sides,

⇒ \( {{{(x - 2)}^2} + {y^2}} = 16 +(x + 2)^2+ {y^2}-8 \sqrt {{{(x + 2)}^2} + {y^2}} \)

⇒ \(x^2-4x+4+y^2=16+x^2+4x+4+y^2-8 \sqrt {{{(x + 2)}^2} + {y^2}} \)

⇒ \(16 +8x= 8 \sqrt {{{(x + 2)}^2} + {y^2}} \)

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).

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 + 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 – 8x – 36y + 4 = 0

By comparing the given equation with Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, we get

⇒ A = 4, B = 0, C = 9, D = - 8, E = - 36 and F = 4

Here, we can see that, B = 0, A ≠ C and AC = 36 > 0

As we know that, if B = 0, A ≠ C and AC > 0 then the non-degenerate equation represents an ellipse

Hence, option B is the correct answer.

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 at the same time.

The above given equation represents a non-degenerate conics whose nature is given below in the table:

CALCULATION:

Given: x2 - 6x - 3y + 21 = 0

By comparing the given equation with Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, we get

⇒ A = 1, B = 0, C = 0, D = - 6, E = - 3 and F = 21

Here, we can see that, B = 0 and C = 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 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)

• \(\rm r=\sqrt{{{x}^{2}}+{{y}^{2}}}\)
• \(\rm \theta ={{\tan }^{-1}}\frac{y}{x}\)

Calculation:

\(\rm \frac{1}{r} = \frac{1}{3} + \frac{2}{3} \cos \theta \)

⇒ 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: \(\rm \frac{l}{r} =1 + e \cos θ \)

Given equation is \(\rm \frac{1}{r} = \frac{1}{16} + \frac{6}{16} cos θ \) or \(\rm \frac{16}{r} =1 + 6 \cos θ \) 

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 \(\frac{x}{1} = \frac{y}{{ - 2}} = \frac{z}{3}\) & \(\frac{x}{3} = \frac{y}{{ - 1}} = \frac{z}{1}\)  

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)

\(⇒ \frac{f}{-9+6} =\frac{g}{2-18} = \frac{h}{-18+3} = k \)

⇒ 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 cos θ  − b sin θ          ....(i) 

y = a sin θ  + b cos θ         ....(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.

Hence Option 4 is correct.

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

Concept used:

The general equation of a sphere is (x - a)² + (y - b)² + (z - c)² = r², where (a, b, c) represents the center of the sphere, 'r' represents the radius, and x, y, and z are the coordinates of the points on the surface of the sphere.

Equation of sphere is

x² + y² + z² + 2ux + 2vy + 2wz + d = 0

also it passes through (1, 0, 0), (0, 1, 0) and (0, 0, 1)

when the sphere passes through (1, 0, 0) we get

x² + y² + z² + 2ux + 2vy + 2wz + d = 0

1² + 0 + 0 + 2u + 0 + 0 + d = 0

1 + 2u + d = 0

2u + d = -1

When the sphere passes through (0, 1, 0) we get

x² + y² + z² + 2ux + 2vy + 2wz + d = 0

0 + 1² + 0 + 0 + 2v + 0 + d = 0

1 + 2v + d = 0

2v + d = -1

Similarly, when the sphere passes through (0, 0, 1) we get

x² + y² + z² + 2ux + 2vy + 2wz + d = 0

0 + 0 + 1 + 0 + 0 + 2w + d = 0

1 + 2w + d = 0

2w + d = -1

Cleary we can see that all the equation depends on the diameter, so there can be infinite no of sphere passing through these points.

Given:

A(3, 4, 5) and B(-1, 3, -7) are two points.

Formula Used:

Distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 } \)

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

⇒ PA = \(\sqrt{(3 - x)^2 + (4 - y)^2 + (5 - z)^2} \)

Squaring on both sides,

⇒ (PA)² = \((\sqrt{(3 - x)^2 + (4 - y)^2 + (5 - z)^2})^2 \)

⇒ (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

⇒ PB = \(\sqrt{(-1 -x)^2 + (3 - y)^2 + (-7 - z)^2} \)

Squaring on both sides,

⇒ (PB)² = \((\sqrt{(-1 -x)^2 + (3 - y)^2 + (-7 - z)^2})^2 \)

⇒ (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