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