Question:
If the angle between the lines given by the equation \(x^2 -3xy + dy^2 + 3x - 5y + 2=0\); \(d > 0\), is \(tan^{-1} (\frac{1}{a})\) then the value of d is?
If the angle between the lines given by the equation \(x^2 -3xy + dy^2 + 3x - 5y + 2=0\); \(d > 0\), is \(tan^{-1} (\frac{1}{a})\) then the value of d is?
Updated On: May 20, 2023
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Solution and Explanation
Given the equation \(x^2 - 3xy + dy^2 + 3x - 5y + 2 = 0\) and \(d > 0\) is \(tan^{-1}(\frac{1}{a})\),
we can determine the value of d as follows: By comparing the equation with the general form of a conic section, \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\), we can see that \(A = 1\), \(B = -3\), \(C = d\), \(D = 3\), \(E = -5\), and \(F = 2\).
For an ellipse, \(B^2 - 4AC > 0\).
Substituting the values, we have \((-3)^2 - 4(1)(d) > 0. 9 - 4d > 0 -4d > -9d < \frac{9}{4}\)
Since \(d > 0\), the maximum possible value for d is \(\frac{9}{4}\).
In brief, the value of d is less than \(\frac{9}{4}\).
we can determine the value of d as follows: By comparing the equation with the general form of a conic section, \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\), we can see that \(A = 1\), \(B = -3\), \(C = d\), \(D = 3\), \(E = -5\), and \(F = 2\).
For an ellipse, \(B^2 - 4AC > 0\).
Substituting the values, we have \((-3)^2 - 4(1)(d) > 0. 9 - 4d > 0 -4d > -9d < \frac{9}{4}\)
Since \(d > 0\), the maximum possible value for d is \(\frac{9}{4}\).
In brief, the value of d is less than \(\frac{9}{4}\).
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Concepts Used:
Angle between Two Lines
The two straight lines, whenever intersects, form two sets of angles. The angles so formed after the intersection are;
- A pair of acute angle
- Another pair of an obtuse angle
The absolute values of angles created depend on the slopes of the intersecting lines.
It is also worth taking note, that the angle so formed by the intersection of two lines cannot be calculated if any of the lines is parallel to the y-axis as the slope of a line parallel to the y-axis is an indeterminate.
Read More: Angle Between Two Lines