Find the abscissa of the point on the curve 3y = 6x − 5x³, the normal at which passes through origin
Answers
x1 = 1 is abscissa on the curve 3y = 6x - 5x³ , the normal at which passes through origin
it is given that, normal at curve 3y = 6x - 5x³ passes through origin.
differentiating with respect to x,
3 dy/dx = 6 - 15x²
⇒dy/dx = 2 - 5x²
slope of tangent = 2 - 5x²
so, the slope of normal to the curve = -1/slope of tangent
= -1/(2 - 5x²)
let point on curve is (x1, y1)
then, slope of normal is -1/(2 - 5x1²)
as it is given that equation of normal is passing through origin.
so, -1/(2 - 5x1²) = (y1 - 0)/(x1 - 0) = y1/x1
⇒-1/(2 - 5x1²) = (6x1 - 5x1³)/3x1 = (6 - 5x1²)/3
Putting, x1 = 1
LHS = -1/-3 = 1/3
RHS = 1/3
so, x1 = 1 is abscissa on the curve 3y = 6x - 5x³ , the normal at which passes through origin.
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