Question

Find the angle of intersection of the curves y² = 4ax and x² = 4by.

Answer 1

Given that y² = 4ax  .....(i) and x² = 4by  .....(ii)

Solving (i) and (ii), we get

`(x^2/(4"b"))^2` = 4ax

⇒ x⁴ = 64 ab²x

or x(x³ – 64 ab²) = 0

⇒ x = 0, x = `4"a"^(1/3) "b"^(2/3)`

Therefore, the points of intersection are (0, 0) and `(4"a"^(1/3) "b"^(2/3), 4"a"^(2/3)"b"^(1/3))`.

Again, y² = 4ax

⇒ `"dy"/"dx" = (4"a")/"dx" = (2"a")/y` and x² = 4by

⇒ `"dy"/"dx" = (2x)/(4"b") = x/(2"b")`

Therefore, at (0, 0) the tangent to the curve y² = 4ax is parallel to y-axis and tangent to the curve x² = 4by is parallel to x-axis.

⇒  Angle between curves = `pi/2`

At `(4"a"^(1/3)"b"^(2/3), 4"a"^(2/3)"b"^(1/3))`, m₁  ......(Slope of the tangent to the curve (i))

= `2("a"/"b")^(1/3)`

= `(2"a")/(4"a"^(2/3)"b"^(1/3))`

= `1/2("a"/"b")^(1/3)`, m₂  ....(Slope of the tangent to the curve (ii))

= `(4"a"^(1/3)"b"^(2/3))/(2"b")`

= `2("a"/"b")^(1/3)`

Therefore, tan θ = `|("m"_2 - "m"_3)/(1 + "m"_1 "m"_2)|`

= `|(2("a"/"b")^(1/3) - 1/2("a"/"b")^(1/3))/(1 + 2("a"/"b")^(1/3)  1/2("a"/"b")^(1/3))|`

= `(3"a"^(1/3) . "b"^(1/3))/(2("a"^(2/3) + "b"^(2/3))`

Hence, θ = `tan^-1((3"a"^(1/3) . "b"^(1/3))/(2("a"^(2/3) + "b"^(2/3))))`