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.