Question
The curve $$\frac{{{x^n}}}{{{a^n}}} + \frac{{{y^n}}}{{{b^n}}} = 2$$ touches the line $$\frac{x}{a} + \frac{y}{b} = 2$$ at the point :
A.
$$\left( {b,\,a} \right)$$
B.
$$\left( {a,\,b} \right)$$
C.
$$\left( {1,\,1} \right)$$
D.
$$\left( {\frac{1}{b},\,\frac{1}{a}} \right)$$
Answer :
$$\left( {a,\,b} \right)$$
Solution :
Differentiating w.r.t.$$x,\,\frac{{n{x^{n - 1}}}}{{{a^n}}} + \frac{{n{y^{n - 1}}}}{{{b^n}}}.\frac{{dy}}{{dx}} = 0$$
$${\text{or }}\frac{{dy}}{{dx}} = - \frac{{{b^n}{x^{n - 1}}}}{{{a^n}{y^{n - 1}}}}$$
$$\therefore $$ The equation of the tangent at $$\left( {\alpha ,\,\beta } \right)$$ is
$$\eqalign{
& y - \beta = - \frac{{{b^n}{\alpha ^{n - 1}}}}{{{a^n}{\beta ^{n - 1}}}}\left( {x - \alpha } \right) \cr
& {\text{or }}{a^n}{\beta ^{n - 1}}y + {b^n}{\alpha ^{n - 1}}x = {a^n}{\beta ^n} + {b^n}{\alpha ^n} \cr} $$
But $$\left( {\alpha ,\,\beta } \right)$$ is on the curve.
$$\therefore \,{a^n}{\beta ^{n - 1}}y + {b^n}{\alpha ^{n - 1}}x = {a^n}{b^n}.2......(1)$$
The line touches the curve at $$\left( {\alpha ,\,\beta } \right)$$ if (1) and $$\frac{x}{a} + \frac{y}{b} = 2$$ are identical.
$$\eqalign{
& \therefore \frac{{{b^n}{\alpha ^{n - 1}}}}{{\frac{1}{a}}} = \frac{{{a^n}{\beta ^{n - 1}}}}{{\frac{1}{b}}} = \frac{{2{a^n}{b^n}}}{2} \cr
& \Rightarrow {\alpha ^{n - 1}} = {a^{n - 1}},\,{\beta ^{n - 1}} = {b^{n - 1}} \cr
& \therefore \,\left( {\alpha ,\,\beta } \right) = \left( {a,\,b} \right) \cr} $$