Question

If the length of subnormal is equal to the length of sub tangent at any point (3,4) on the curve y=f(x) and the tangent at (3,4) to y=f(x) meets the coordinate axis at A and B, the maximum area of \[\vartriangle OAB.\] where O is origin, is

Answer 1

Hint: Important formulas used to solve such questions:
Length of subtangent- \[y \times \dfrac{{dx}}{{dy}}\]
Length of subnormal- \[y \times \dfrac{{dy}}{{dx}}\]
Equation of tangent passing through the point (a,b) and having slope m is given by- \[(y - a) = m(x - b)\]

Complete step-by-step answer:

Given: Length of subtangent= length of subnormal
\[
   \Rightarrow y \times \dfrac{{dx}}{{dy}} = y \times \dfrac{{dy}}{{dx}} \\
   \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = 1 \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \pm 1 \\
\]
Now, Equation of tangent passing through point (3,4) and having slope 1 is given as:
\[
   \Rightarrow (y - a) = m(x - b) \\
   \Rightarrow (y - 4) = 1(x - 3) \\
   \Rightarrow y = x + 1 \\
\]
So, the coordinates of A and B are (-1,0) and (0,1) respectively.
Area of \[\triangle OAB\]is given as:
\[
   \Rightarrow \dfrac{1}{2} \times OA \times OB \\
   \Rightarrow \dfrac{1}{2} \times 1 \times 1......(O = origin) \\
   \Rightarrow \dfrac{1}{2}sq.units \\
\]
Similarly, equation of tangent passing through point (3,4) and having slope 1 is given as:
\[
   \Rightarrow (y - a) = m(x - b) \\
   \Rightarrow (y - 4) = ( - 1)(x - 3) \\
   \Rightarrow y - 4 = 3 - x \\
   \Rightarrow y + x = 7 \\
\]
So, the coordinates of A and B are (7,0) and (0,7) respectively.
Area of \[\triangle OAB\]is given as:
\[
   \Rightarrow \dfrac{1}{2} \times OA \times OB \\
   \Rightarrow \dfrac{1}{2} \times 7 \times 7......(O = origin) \\
   \Rightarrow \dfrac{{49}}{2}sq.units \\
\]
So, the maximum area of \[\triangle OAB\] is \[\dfrac{{49}}{2}sq.units\].

Note: In above question we have given that the tangent meets coordinate axes means it will meet at abscissa and ordinate axes. The above coordinates can be found by putting once x=0 and then y=0.