Question

If the length of sub-normal is equal to the length of sub-tangent at point (3, 4) on the curve y = f(x) and the tangent at (3, 4) to y = f(x) meets the coordinate axes at a and b, then the maximum area of △oab△oab , what is the value of o?

Answer 1

This is probably the correct procedure

Answer 2

As, given, for curve , y= f(x)

Length of sub-normal =  Length of sub-tangent

Length of sub-tangent= $\frac{y}{y'}$

y'= Slope of tangent

Length of sub-normal =y y'

y y'= $\frac{y}{y'}$

y'²=1

Gives, y'=1 or y'= -1

And,  the tangent at (3, 4) to y = f(x) meets the coordinate axes at a and b.

Equation of tangent passing through (3,4) and slope = 1 or -1 ,is given by

$\frac{y-4}{x-3}=\pm 1$

y-4 = x-3 and y-4= -x +3

x-y=-1 and x +y=7,  are two tangents on the curve f(x), which passes through (3,4).

Tangent Line 1. x-y = -1

it cuts the X axis at a(-1,0) and y axis at b(0,1).

If O is the origin , then Area (Δo ab)=$\frac{1}{2}\times oa \times ob=\frac{1}{2}\times 1 \times 1=\frac{1}{2}$ square units

Tangent Line 2. x +y=7

it cuts the X axis at a(7,0) and y axis at b(0,7).

If O is the origin , then Area (Δo ab)=$\frac{1}{2}\times oa \times ob=\frac{1}{2}\times 7 \times 7=\frac{49}{2}$ square units