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Subtangent and Subnormal

Sub tangent and Subnormal

subtangent

In the figure given above PT' is tangent to the curve at point P of the curve and PN'  is normal. Point T is on x axis where tangent intersects it and point N is on x axis where normal PN'  meets it.

TM'  is called sub-tangent and MN'  is called Subnormal.

From figure

        TN'  = PM'  cot a

        = (f(a))/(f' (a)) (f'(a) ≠ 0)

        Also, MN'  = PM' . tan a

       = f(a) . f'(a)

Illustration:

        Show that the sum of the intercepts of the tangents to the curve 
        √x + √y = √a on the coordinate axis is constant.

Solution:

        The equation of the curve is √x + √y = √a.

        Differentiating with respect to x, we have

        1/2 x-1/2 + 1/2 y-1/2 dy/dx = 0 => dy/dx=-√(y/x).

        The equation of the tangent at (x1, y1) is

        y - y1 = √(y1/x1 ).

The intercept on the x-axis = √(ax1 ).

The intercept on the y-axis = √(ay1 ).

Sum of the intercepts = √(ax1 ) + √(ay1 ) = √a (√(ax1) + √(ay1))

                             = a, which is a constant.

Illustration:

The curve = ax3 + bx2 + cx + 5 touches the x-axis at P(-2, 0) and cuts the y-axis at a point Q where its gradient is 3. Find a, b, c.

Solution:

       Slope of the tangent to the curve at (x1, y1) is

       [dy/dx](x1,y1) = 3ax12 + 2bx1 + c

       The point Q is (0, 5).

Since the curve passes through (-2, 0),

-8a + 4b - 2c + 5 = 0.                                                         ... (1)

Since the slope of the tangent at (-2, 0) is 0,

12a - 4b + c = 0                                                                 ... (2)

Since the slope of the tangent at (0, 5) is 3,

c = 3.                                                                                ... (3)

From (1), (2) and (3), a = -1/2,b = -3/4, c = 3.

Illustration:

Find the equation of tangent and normal to the curve x2 = 8y + 6 at the point (0, - 5/4)

Solution:

Equation of the curve is

                        x2 = 8y + 6

                        y = (x2-6)/8

Slope of the tangent to the curve at (0, -5/4) is dy/dx at this particular point

                dy/dx=2x/8=x/4

                at (0, -5/4) dy/dx = 0

                equation of tangent at (0, -5/4)

                y + y1 = m(x - x1)

                y + 5/4 = 0(x - 0)

                y + 5/4 = 0

                normal is perpendicular to tangent, so slope of normal = - 1/(dy/dx).

Since dy/dx = 0 hence the normal will be parallel to y axis and will pass through (0, - 5/4), hence equation of normal at (0, -5/4) is 

                               x = 0

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USE CODE: askschfd09