The asymptotes of the curve 2x2 + 5xy + 2y2 + 4x + 5y = 0 are given by...
The hyperbola is given by
2x2 + 5xy + 2y2 + 4x + 5y = 0 ............................(i)
Since the equation of hyperbola will differ from equation of asymptote by a constant. So, equation of asymptote is
2x2 + 5xy + 2y3 + 4x + 5y + λ = 0 ......................(ii)
If (ii) represents 2 straight lines, we must have
abc + 2fgh - af2- bg2 - ch2 = 0
or 22λ + 2.5 / 2.4 / 2.5 / 2 - 2.25 / 4 - 2.4 - λ (25 / 4) = 0
or 9λ = 18 ∴ λ = 2
2x2 + 5xy + 2y3 + 4x + 5y + 2 = 0
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The asymptotes of the curve 2x2 + 5xy + 2y2 + 4x + 5y = 0 are given by...
Understanding the Curve
The given equation of the curve is:
2x^2 + 5xy + 2y^2 + 4x + 5y = 0.
This is a second-degree polynomial, and its general form indicates that it might represent a conic section. To find the asymptotes, we need to analyze the leading coefficients and their interactions.
Identifying the Asymptotes
To derive the asymptotes, we can consider the homogeneous part of the equation:
2x^2 + 5xy + 2y^2 = 0.
We can factor this equation to identify lines that describe the asymptotic behavior as x and y approach infinity.
Factoring the Homogeneous Part
To factor, we can treat the equation as a quadratic in y:
2y^2 + 5xy + 2x^2 = 0.
Using the quadratic formula, we can find the roots which represent the slopes of the asymptotes:
y = (-5x ± √(25x^2 - 16x^2))/(4).
This simplifies to:
y = (-5x ± 3x)/(4).
This leads to two asymptotes:
y = -x/2 and y = -2x.
Deriving the Asymptotic Equation
To get the explicit equation of the asymptotes, we need to eliminate the constant term from the original equation. The asymptotes can be represented by the modified equation:
2x^2 + 5xy + 2y^2 + 4x + 5y + 2 = 0.
This is option 'A', indicating that the asymptotes are represented by this equation, confirming the correct answer.
Conclusion
Thus, the curve's asymptotes are effectively captured by option 'A', providing a structured understanding of its behavior at infinity.