Question

If $\left( \frac { a } { b } \right) ^ { 1 / 3 } + \left( \frac { b } { a } \right) ^ { 1 / 3 } = \frac { \sqrt { 3 } } { 2 }$, then the angle of intersection of the parabola $y ^ { 2 } = 4 a x$ and $x ^ { 2 } = 4 b y$ at a point other than the origin is

• A. $\pi / 4$
• B. $\pi / 3$
• C. $\pi / 2$
• D. None

Answer 1

Answer: (B)

$\pi / 3$

Answer 2

Given parabolas are $y ^ { 2 } = 4 a x$ ....(i) and $x ^ { 2 } = 4 b y$ .....(ii)

These meet at the points (0, 0), $\left( 4 a ^ { 1 / 3 } b ^ { 2 / 3 } , 4 a ^ { 2 / 3 } b ^ { 1 / 3 } \right)$

Tangent to (i) at $\left( 4 a ^ { 1 / 3 } b ^ { 2 / 3 } , 4 a ^ { 2 / 3 } b ^ { 1 / 3 } \right)$is

Slope of the tangent

$= \frac { a ^ { 1 / 3 } } { 2 b ^ { 1 / 3 } }$

Tangent to (ii) at $\left( 4 a ^ { 1 / 3 } b ^ { 2 / 3 } , 4 a ^ { 2 / 3 } b ^ { 1 / 3 } \right)$is

Slope of the tangent $\left( m _ { 2 } \right)$ $= \frac { 2 a ^ { 1 / 3 } } { b ^ { 1 / 3 } }$

If $\theta$ is the angle between the two tangents, then

$= \frac { 3 } { 2 } \cdot \frac { 1 } { \left( \frac { a } { b } \right) ^ { 1 / 3 } + \left( \frac { b } { a } \right) ^ { 1 / 3 } } = \frac { 3 } { 2 } \cdot \frac { 1 } { \frac { \sqrt { 3 } } { 2 } } = \sqrt { 3 }$; ∴$\theta = 60 ^ { \circ } = \frac { \pi } { 3 }$