Find the equations of the tangent and normal to the parabola y² = 4ax at the point (at², 2at).
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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
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
plz mark me as BRILLIANT ❤️...
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