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Question: There is a point \(\left( {p,q} \right)\) on the graph of \(f\left( x \right) = {x^2}\)and a point \...

There is a point (p,q)\left( {p,q} \right) on the graph of f(x)=x2f\left( x \right) = {x^2}and a point (r,s)\left( {r,s} \right) on the graph of g(x)=−8xg\left( x \right) = \dfrac{{ - 8}}{x} where p>0,q>0p > 0,q > 0. If line through (p,q)\left( {p,q} \right) and (r,s)\left( {r,s} \right) is also tangent to both the curves at these points respectively, then find the value of (p+r)\left( {p + r} \right).

Explanation

Solution

In order to get the value of (p+r)\left( {p + r} \right), take the two functions separately with their points, put the points in the functions, find their derivative, and compare the values with the two functions as its given that the lines passing through the points is tangent to both the curves given. Simplify the values one by one and get the answer.

Formula used:

  1. f′(p)f'\left( p \right) is the derivative of function f(p)f\left( p \right), with respect to pp.
  2. f′(xn)=nxn−1f'\left( {{x^n}} \right) = n{x^{n - 1}}
  3. f′(1x)=−1x2f'\left( {\dfrac{1}{x}} \right) = \dfrac{{ - 1}}{{{x^2}}}
  4. Slope of tangent for points (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right) is y2−y1x2−x1\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}

Complete step by step solution:
We are given two functions, one is f(x)=x2f\left( x \right) = {x^2} and another is g(x)=−8xg\left( x \right) = \dfrac{{ - 8}}{x}.
Since, a point (p,q)\left( {p,q} \right) lies on the graph of the first function that is f(x)=x2f\left( x \right) = {x^2}. So, replacing the co-ordinates of the function with the given point and we get:
f(p)=p2 q=p2  f\left( p \right) = {p^2} \\\ q = {p^2} \\\
Similarly, a point (r,s)\left( {r,s} \right) lies on the graph of the second function that is g(x)=−8xg\left( x \right) = \dfrac{{ - 8}}{x}. So, replacing the co-ordinates of the function with the given point and we get:
g(r)=−8r s=−8r  g\left( r \right) = \dfrac{{ - 8}}{r} \\\ s = \dfrac{{ - 8}}{r} \\\

It is given that the line through (p,q)\left( {p,q} \right) and (r,s)\left( {r,s} \right) is also tangent to both the curves of the functions at these points respectively, that gives:
Slope of tangent =f′(p)=g′(r)== f'\left( p \right) = g'\left( r \right) =Slope of (p,q)\left( {p,q} \right)to (r,s)\left( {r,s} \right).
Solving these terms separately:
f(p)=p2 =>f′(p)=2p  f\left( p \right) = {p^2} \\\ = > f'\left( p \right) = 2p \\\
Similarly,
g(r)=−8r g′(r)=8r2  g\left( r \right) = \dfrac{{ - 8}}{r} \\\ g'\left( r \right) = \dfrac{8}{{{r^2}}} \\\

And, Slope of (p,q)\left( {p,q} \right)to (r,s)=s−qr−p\left( {r,s} \right) = \dfrac{{s - q}}{{r - p}}.
Solving the terms for variables by taking two operands at once:
f′(p)=g′(r) 2p=8r2 p=4r2  f'\left( p \right) = g'\left( r \right) \\\ 2p = \dfrac{8}{{{r^2}}} \\\ p = \dfrac{4}{{{r^2}}} \\\
Next, two operands:
f′(p)=f'\left( p \right) = Slope of (p,q)\left( {p,q} \right)to (r,s)=s−qr−p\left( {r,s} \right) = \dfrac{{s - q}}{{r - p}}
f′(p)=s−qr−p 2p=s−qr−p  f'\left( p \right) = \dfrac{{s - q}}{{r - p}} \\\ 2p = \dfrac{{s - q}}{{r - p}} \\\
Substituting, q=p2q = {p^2} and s=−8rs = \dfrac{{ - 8}}{r} in the above equation and we get:
2p=s−qr−p 2p=−8r−p2r−p  2p = \dfrac{{s - q}}{{r - p}} \\\ 2p = \dfrac{{\dfrac{{ - 8}}{r} - {p^2}}}{{r - p}} \\\
On further solving, we get:
2p=−8r−p2r−p 2p(r−p)=−8r−p2 2pr−2p2=−8r−p2 2pr+8r=2p2−p2 2pr+8r=p2  2p = \dfrac{{\dfrac{{ - 8}}{r} - {p^2}}}{{r - p}} \\\ 2p\left( {r - p} \right) = \dfrac{{ - 8}}{r} - {p^2} \\\ 2pr - 2{p^2} = \dfrac{{ - 8}}{r} - {p^2} \\\ 2pr + \dfrac{8}{r} = 2{p^2} - {p^2} \\\ 2pr + \dfrac{8}{r} = {p^2} \\\
Substituting the value of p=4r2p = \dfrac{4}{{{r^2}}} above and we get:
2pr+8r=p2 2.4r2.r+8r=(4r2)2 8r+8r=16r4 16r=16r4 r3=1 =>r=1  2pr + \dfrac{8}{r} = {p^2} \\\ 2.\dfrac{4}{{{r^2}}}.r + \dfrac{8}{r} = {\left( {\dfrac{4}{{{r^2}}}} \right)^2} \\\ \dfrac{8}{r} + \dfrac{8}{r} = \dfrac{{16}}{{{r^4}}} \\\ \dfrac{{16}}{r} = \dfrac{{16}}{{{r^4}}} \\\ {r^3} = 1 \\\ = > r = 1 \\\
Substituting value of rr in p=4r2p = \dfrac{4}{{{r^2}}} and we get:
p=4r2=41=4p = \dfrac{4}{{{r^2}}} = \dfrac{4}{1} = 4
Hence, p=4p = 4 and r=1r = 1, that implies p+r=4+1=5p + r = 4 + 1 = 5

Therefore, the value of (p+r)=5\left( {p + r} \right) = 5.

Note:

It’s important to evaluate the slopes by taking the operands two at once to ease the solving process, otherwise it would make it complicated to solve.
Always solve step by step rather than solving at once.
Cross check your answer once, before coming to a conclusion.