Question
If a tangent to the parabola 4y2 = x makes an angle of 60º with the x- axis, then its point of contact is-
Hint:
find out the values of m and a and substitute their values.
The correct answer is:
= (1/48, 1/8√3)
m= tan(60) = 1.732
a= 1/16
point of contact = (a/m2, 2a/m)
= (1/48, 1/8√3)
point of contact = (a/m2, 2a/m)
slope = tan ( angle made with the x axis)
this gives us the value of m
y^2 = x/4 = 4(1/16)x
a = 1/16
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