Question

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

Answer 1

HELLO DEAR,

given curves are y² = 4ax at point (at² , 2at).

now, 2y * dy/dx = 4a

dy/dx = 2a/y

the slope of tangent at point at point (at² , 2at) is

$\bold{frac{dy}{dx}_{at^2,2at} = 2a/2at = 1/t}$

Equation of the tangent at (x1 , y1) where slope is m is given by y − y1 = m(x−x1)
where, m = dy/dx
hence, the equation of tangent is:

y - 2at = dy/dx(x - at²)

y - 2at = 1/t(x - at²)

yt - 2at² = x - at²

x - yt + at² = 0


Equation of the normal at (x1 , y1) where slope is m is given by y − y1 = -1/m(x−x1)
where, m = dy/dx ,
hence, equation of normao is:

y - 2at = -1/(1/t)(x - at²)

y - 2at = -xt + at³

y + xt = at³ + 2at.


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

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