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Question
Find the angle of intersection of the curves y2 = 4ax and x2 = 4by.
Solution
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))))`
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