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?

Answer 1

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 $\frac{9}{4}$.