General Equation of Conics MCQ Quiz in मल्याळम - Objective Question with Answer for General Equation of Conics - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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

Concept Used:-

If the standard equation \(a x^2+2 h x y+b y^2+2 g x+2 f y+c=0\) represents a pair of straight line, then it follow the following condition,

\(a b c+2 f g h-a f^2-b g^2-c h^2=0\)

Explanation:-

Given equation of pair of straight lines is,

2x2 + 7xy + 3y2 + 8x + 14y + λ = 0 

So, from the above concept,

\(6 \lambda+2(7)(4)\left(\frac{7}{2}\right)-2(7)^2-3(4)^2-\lambda\left(\frac{7}{2}\right)^2=0\)

\(\begin{aligned} & \Rightarrow 6 \lambda+196-98-48-\frac{49 \lambda}{4}=0 \\ & \Rightarrow \frac{49 \lambda}{4}-6 \lambda=196-146 \\ & \Rightarrow \frac{25 \lambda}{4}=50 \\ & \Rightarrow \lambda=\frac{50 \times 4}{25}\\ & \Rightarrow \lambda=8 \end{aligned}\)

Hence, the value of λ is 8. Hence, the correct option is 4.

Concept:

The distance of any point of the curve from focus is equal to e times

the distance of that point from the directrix, where e is the eccentricity.

Calculation:

Let (x,y) be any point on the conic

⇒ \(\sqrt{(x-1)^2 +(y+1)^2} = e |{x-y +1 \over \sqrt {1^2 +1^2}}|\)

⇒ \(\sqrt{(x-1)^2 +(y+1)^2} = \sqrt{2} |{x-y +1 \over \sqrt {1^2 +1^2}}|\)

⇒ \({(x-1)^2 +(y+1)^2} = {2} |{x-y +1 \over \sqrt {2}}|^2\)

⇒ x² - 2x + 1 + y² + 2y + 1 = (x - y + 1)²

⇒ x2 - 2x + 1 + y2 + 2y + 1 = x2 + y2 + 1 - 2xy + 2x - 2y

⇒ 2xy - 4x + 4y + 1 = 0

∴ The correct answer is option (3).

Calculation:

Given equation is  3x2 + 2xy + 3y2 - 16x + 20 = 0 

Comparing the equation with the standard form, ax² + 2hxy + by² + 2gx + 2fy + c = 0.

⇒ a = 3, h = 1, b = 3, g = - 8, f = 0, c = 20.

Δ = abc + 2fgh - af² - bg² - ch²

⇒ Δ = 3 × 3 × 20 + 2 × 0 ×  gh - a × 0 - 3 × (-8)² - 20 × 1

⇒ Δ = 180 - 192 - 20 = - 32

⇒ Δ < 0.

Discriminant = h² - ab

⇒ Discriminant = 1 - 3 × 3 = - 8

∵ Δ and discriminant are less than 0; the curve is either a circle or an ellipse.

Consider \(\rm\frac{\Delta }{a+b}\) = \(\frac{-32}{3+3}\) = \(\frac{-32}{6} < 0\)

∴ The curve is an ellipse.

Concept:

The general second order equation:

Ax2 + 2Hxy + By2 + 2Gx + 2Fy + C = 0, represents

Circle: a = b and h = 0

Pair of lines: ABC + 2FGH - AF2 - BG2 - CH2 = 0

Calculation:

Given that,

ax2 + by2 + hx + hy = 0 and h ≠ 0 

On comparing it with general equation mentioned above

A = a, B = b, H = 0, G = h/2, F = h/2 & C = 0

We know that, for pair of straight line,

ABC + 2FGH - AF2 - BG2 - CH2 = 0

⇒ 0 + 0 - a(h/2)² - b(h/2)² - 0 = 0

⇒ (a + b)h²/4 = 0

⇒ a + b = 0