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Mathematics
The slope of the normal at the point (at2, 2at) of the parabola y2 = 4ax is
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Q. The slope of the normal at the point $(at^2, 2at)$ of the parabola $ y^2 = 4ax$ is
Conic Sections
A
$\frac{1}{t}$
23%
B
t
18%
C
-t
40%
D
$-\frac{1}{t}$
18%
Solution:
Tangent at $'t'$ is $ty= x+at^{2} $
Slope of the tangent $= \frac{1}{t} $
$ \therefore $ slope of normal $= -t $