Question

The three tangents to the parabola y² = 4ax, are such that the tangents of their inclination to the axis of the parabola are in H.P. with common difference of its corresponding A.P is d, form a triangle whose area is-

• A. ad
• B. $\frac{a}{d}$
• C. $\frac{ad}{2}$
• D. a²d³

Answer 1

Answer: (D)

a²d³

Answer 2

Let the slopes of tangents are m₁, m₂, m₃.

Hence tangents are y = m₁x + $\frac{a}{m_{1}}$,

y = m₂x + $\frac{a}{m_{2}}$and y = m₃x +$\frac{a}{m_{3}}$

also slopes are in H.P.

So $\frac{1}{m_{2}}$– $\frac{1}{m_{1}}$= $\frac{1}{m_{3}}$–$\frac{1}{m_{2}}$= d

coordinates of the angular points of triangle

$\left( \frac{a}{m_{1}m_{2}},a\left( \frac{1}{m_{1}} + \frac{1}{m_{2}} \right) \right)$, $\left( \frac{a}{m_{2}m_{3}},a\left( \frac{1}{m_{2}} + \frac{1}{m_{3}} \right) \right)$,

$\left( \frac{a}{m_{3}m_{1}},a\left( \frac{1}{m_{3}} + \frac{1}{m_{1}} \right) \right)$

D = $\frac { \mathrm { a } ^ { 2 } } { 2 }$ $\left( \frac{1}{m_{1}} - \frac{1}{m_{2}} \right)$$\left( \frac{1}{m_{3}} - \frac{1}{m_{1}} \right)$

D = $\frac{a^{2}}{2}$ (–d) (–d) (2d) ̃ D = a²d³