Question

If the tangent at $ \left( {{x}_{1}},{{y}_{1}} \right) $ to the curve $ {{x}^{3}}+{{y}^{3}}={{a}^{3}} $ meets the curve again at $ \left( {{x}_{2}},{{y}_{2}} \right) $ , then
A. $ \dfrac{{{x}_{2}}}{{{x}_{1}}}+\dfrac{{{y}_{2}}}{{{y}_{1}}}=-1 $
B. $ \dfrac{{{x}_{2}}}{{{y}_{1}}}+\dfrac{{{x}_{1}}}{{{y}_{2}}}=-1 $
C. $ \dfrac{{{x}_{1}}}{{{x}_{2}}}+\dfrac{{{y}_{1}}}{{{y}_{2}}}=-1 $
D. $ \dfrac{{{x}_{2}}}{{{x}_{1}}}+\dfrac{{{y}_{2}}}{{{y}_{1}}}=1 $

Answer 1

Hint: We first find the slopes for the curve $ {{x}^{3}}+{{y}^{3}}={{a}^{3}} $ at $ \left( {{x}_{1}},{{y}_{1}} \right) $ which is same to the value of slope of the line joining the points $ \left( {{x}_{1}},{{y}_{1}} \right) $ and $ \left( {{x}_{2}},{{y}_{2}} \right) $ . We put the points $ \left( {{x}_{1}},{{y}_{1}} \right) $ and $ \left( {{x}_{2}},{{y}_{2}} \right) $ in the curve $ {{x}^{3}}+{{y}^{3}}={{a}^{3}} $ . We solve the equation from the relation.

Complete step-by-step answer:
The tangent at $ \left( {{x}_{1}},{{y}_{1}} \right) $ to the curve $ {{x}^{3}}+{{y}^{3}}={{a}^{3}} $ meets the curve again at $ \left( {{x}_{2}},{{y}_{2}} \right) $ .
The slope to the curve $ {{x}^{3}}+{{y}^{3}}={{a}^{3}} $ at $ \left( {{x}_{1}},{{y}_{1}} \right) $ will be same to the value of slope of the line joining the points $ \left( {{x}_{1}},{{y}_{1}} \right) $ and $ \left( {{x}_{2}},{{y}_{2}} \right) $ .
We first find slope to the curve $ {{x}^{3}}+{{y}^{3}}={{a}^{3}} $ at $ \left( {{x}_{1}},{{y}_{1}} \right) $ . We find derivatives of the curve $ {{x}^{3}}+{{y}^{3}}={{a}^{3}} $ .
 $
   \dfrac{d}{dx}\left( {{x}^{3}}+{{y}^{3}} \right)=\dfrac{d}{dx}\left( {{a}^{3}} \right) \\
  \Rightarrow 3{{x}^{2}}+3{{y}^{2}}\dfrac{dy}{dx}=0 \\
 $
Simplifying we get $ \dfrac{dy}{dx}=-\dfrac{{{x}^{2}}}{{{y}^{2}}} $ . At point $ \left( {{x}_{1}},{{y}_{1}} \right) $ will be $ {{\left[ \dfrac{dy}{dx} \right]}_{\left( {{x}_{1}},{{y}_{1}} \right)}}=-\dfrac{{{x}_{1}}^{2}}{{{y}_{1}}^{2}} $ .
Now the value of slope of the line joining the points $ \left( {{x}_{1}},{{y}_{1}} \right) $ and $ \left( {{x}_{2}},{{y}_{2}} \right) $ is $ \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} $ .
Considering the relation, we get $ \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=-\dfrac{{{x}_{1}}^{2}}{{{y}_{1}}^{2}} $ . Now we have to simplify.
Both $ \left( {{x}_{1}},{{y}_{1}} \right) $ and $ \left( {{x}_{2}},{{y}_{2}} \right) $ goes through curve $ {{x}^{3}}+{{y}^{3}}={{a}^{3}} $ .
So, $ {{x}_{1}}^{3}+{{y}_{1}}^{3}={{a}^{3}} $ and $ {{x}_{2}}^{3}+{{y}_{2}}^{3}={{a}^{3}} $ .
Subtracting we get $ {{x}_{1}}^{3}-{{x}_{2}}^{3}={{y}_{2}}^{3}-{{y}_{1}}^{3} $ which gives \[\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=-\dfrac{{{x}_{2}}^{2}+{{x}_{1}}^{2}+{{x}_{1}}{{x}_{2}}}{{{y}_{2}}^{2}+{{y}_{1}}^{2}+{{y}_{1}}{{y}_{2}}}\].
So, \[\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=-\dfrac{{{x}_{1}}^{2}}{{{y}_{1}}^{2}}=-\dfrac{{{x}_{2}}^{2}+{{x}_{1}}^{2}+{{x}_{1}}{{x}_{2}}}{{{y}_{2}}^{2}+{{y}_{1}}^{2}+{{y}_{1}}{{y}_{2}}}\].
Simplifying we get
\[
   {{x}_{1}}^{2}{{y}_{2}}^{2}+{{x}_{1}}^{2}{{y}_{1}}^{2}+{{x}_{1}}^{2}{{y}_{1}}{{y}_{2}}={{x}_{2}}^{2}{{y}_{1}}^{2}+{{x}_{1}}^{2}{{y}_{1}}^{2}+{{x}_{1}}{{x}_{2}}{{y}_{1}}^{2} \\
  \Rightarrow {{x}_{1}}^{2}{{y}_{2}}^{2}-{{x}_{2}}^{2}{{y}_{1}}^{2}={{x}_{1}}{{x}_{2}}{{y}_{1}}^{2}-{{y}_{1}}{{y}_{2}}{{x}_{1}}^{2} \\
  \Rightarrow {{\left( {{x}_{1}}{{y}_{2}} \right)}^{2}}-{{\left( {{x}_{2}}{{y}_{1}} \right)}^{2}}={{x}_{1}}{{y}_{1}}\left( {{x}_{2}}{{y}_{1}}-{{y}_{2}}{{x}_{1}} \right) \\
\]
Now we use the identity of $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ .
\[
   {{\left( {{x}_{1}}{{y}_{2}} \right)}^{2}}-{{\left( {{x}_{2}}{{y}_{1}} \right)}^{2}}={{x}_{1}}{{y}_{1}}\left( {{x}_{2}}{{y}_{1}}-{{y}_{2}}{{x}_{1}} \right) \\
  \Rightarrow \left( {{x}_{1}}{{y}_{2}}+{{y}_{1}}{{x}_{2}} \right)\left( {{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}} \right)={{x}_{1}}{{y}_{1}}\left( {{x}_{2}}{{y}_{1}}-{{y}_{2}}{{x}_{1}} \right) \\
  \Rightarrow {{x}_{1}}{{y}_{2}}+{{x}_{2}}{{y}_{1}}=-{{x}_{1}}{{y}_{1}} \\
  \Rightarrow \dfrac{{{x}_{2}}}{{{x}_{1}}}+\dfrac{{{y}_{2}}}{{{y}_{1}}}=-1 \\
\]
The correct option is A.
So, the correct answer is “Option A”.

Note: The slope of the curve $ {{x}^{3}}+{{y}^{3}}={{a}^{3}} $ at points $ \left( {{x}_{1}},{{y}_{1}} \right) $ and $ \left( {{x}_{2}},{{y}_{2}} \right) $ will not be same. So, we cannot equate them. We solve the slopes for the equations separately. The slopes are equal as the points $ \left( {{x}_{1}},{{y}_{1}} \right) $ and $ \left( {{x}_{2}},{{y}_{2}} \right) $ lies on the same line.