Question

Angle between a pair of tangents drawn at the end points of the chord y + St = tx + 2 of curve C ∀t∈R, is

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{2}$

Answer 1

Answer: (D)

$\frac{\pi}{2}$

Answer 2

The equation of the chord is given as $y + St = tx + 2$. Rearranging this, we get $y = tx + (2-St)$. If we interpret this as a family of tangents to the parabola $y^2 = 4ax$, then for a tangent $y=mx+c$, the condition is $c = a/m$. Here, $m=t$ and $c=2-St$. Therefore, $2-St = a/t$, which implies $St^2 - 2t + a = 0$. Let $t_1$ and $t_2$ be the roots of this quadratic equation, representing the slopes of the tangents. The angle $\theta$ between the tangents is given by $\tan\theta = \left|\frac{t_1-t_2}{1+t_1t_2}\right|$. From Vieta's formulas, $t_1t_2 = a/S$. For the angle to be constant and independent of $t$, we consider the case where the tangents are perpendicular, i.e., $1+t_1t_2 = 0$. This gives $1+a/S = 0$, or $aS = -1$. In this case, $\tan\theta$ is undefined, implying $\theta = \frac{\pi}{2}$. The discriminant of $St^2 - 2t + a = 0$ is $4-4aS$. If $aS=-1$, the discriminant is $4-4(-1)=8>0$, ensuring two distinct real roots for $t$. Thus, the angle between the tangents is $\frac{\pi}{2}$.