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Q.
If a line $y = 3x + 1$ cuts the parabola $x^2 - 4x - 4y + 20 = 0$ at $A$ and $B$, then the tangent of the angle subtended by line segment $AB$ at the origin is
Conic Sections
Solution:
The joint equation of $OA$ and $OB$ is
$x^2 - 4x (y - 3x) - 4y ( y - 3x) + 20(y - 3x)^2 = 0$
Making the equation of the parabola homogeneous using straight line, we get
$x^2(1 + 12 + 180) - y^2( 4 - 20) - xy ( 4 - 12 + 120) = 0$
or $193x^2 + 16y^2 - 112 xy = 0$
$tan\,\theta = \frac{2\sqrt{h^2-ab}}{a+b}$
$ = \frac{2\sqrt{56^2 - 193 \times 16}}{193 + 16} =\frac{8\sqrt{3}}{209}$