IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

ashgirl

1. for curve y= 4x³-2x⁵, find all pts at which the tangent passes thru origin.

2.find pts on the curve x2+ y2 -2x-3=0 at which tangent r parallel to x- axis?

joyfrancis

slope of tangent at (x1,y1) = y' = 12x₁² - 10x₁⁴

Equation of tangent at a point (x1,y1) to curve: y' (x-x1) = y - y1

i.e 12x₁² - 10x₁⁴( x- x₁) = y-y₁

This passes through the orgin

.: 12x₁² - 10x₁⁴( 0 - x₁) = 0 - y₁........(1)

& y₁ = 4x₁³- 2x₁⁵

Put value of y₁ in (1)

=> 12x₁² - 10x₁⁴( - x₁) = - ( 4x₁³- 2x₁⁵)

x₁= 0,1,-1

.: y1 = 0,2,-2

Hence the points on the curve from where tangents pass through orgin are

(1,2) and (-1,-2)

joyfrancis

q2) Eqn of curve : x2+y2-2x-3 = 0

slope of tangent at any point x1,y1 on the curve is y' i.e dy/dx.

differentiate the eqn

2x + 2yy' - 2 = 0

y' = 1-x/y

.: at a point x1,y1 the slope of tangent is 1-x1/y1

If the tangent is perpendicular to the x axis ,

slope = 0 = 1-x1

.: x1 = 1

and hence y1 = plusminus2

points are (1,2) and (1,-2)