Vikas TU
Last Activity: 8 Years ago
Differntiate and put y also,
y^2=4axydy/dx = 2a
root(ax)dy/dx = a
dy/dx = a/root(ax) = > root(a/x).....................(1)
Differntiate same as
x^2 = 4ay
2x =4dy/dx
x = 2dy/dx
dy/dx = x/2
normal will be:
x/2 * n = -1
n = -2/x …...............(2)
eqn. (1) and eqn. (2) are equal.
therfore,
-2/x = root(a/x)
squaring both sides,
4/x^2 = a/x
4/x^2 – a/x = 0
(4 – ax)/x^2 = 0
x cannot be zero
or
4 – ax = 0
ax = 4
x = 4/a
y = 4/a^2
put these x and y in y^2 = 4ax
u will get,
a^4 = 1
(a^2 + 1) (a^2-1) = 0
a^2 = 1
a = 1 and a = -1
