22) Find the equations of the tangent and normal to the parabola y ^2 = 4 ax at the point (at ^2, 2at).

Answers (1)

Equation of the given curve is
y ^2 = 4 ax

Slope of tangent = 2y\frac{dy}{dx} = 4a \Rightarrow \frac{dy}{dx} = \frac{4a}{2y}
at point (at ^2, 2at).
\frac{dy}{dx}= \frac{4a}{2(2at)} = \frac{4a}{4at} = \frac{1}{t}
Now, the equation of tangent with point (at ^2, 2at). and slope \frac{1}{t} is
y-y_1=m(x-x_1)\\ y-2at=\frac{1}{t}(x-at^2)\\ yt - 2at^2 = x - at^2\\ x-yt +at^2 = 0

We know that
Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = -t
Now, the equation of at point  (at ^2, 2at).  with slope -t
y-y_1=m(x-x_1)\\ y-2at=(-t)(x-at^2)\\ y - 2at = -xt + at^3\\ xt+y -2at -at^3 = 0
 


Hence, the equations of the tangent and normal to the parabola

y ^2 = 4 ax at the point (at ^2, 2at). are
x-yt+at^2=0\ \ \ \ and \ \ \ \ xt+y -2at -at^3 = 0 \ \ respectively

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