Question
The number of tangents to the curve $${x^{\frac{3}{2}}} + {y^{\frac{3}{2}}} = {a^{\frac{3}{2}}},$$ where the tangents are equally inclined to the axes, is :
A.
2
B.
1
C.
0
D.
4
Answer :
1
Solution :
$$\eqalign{
& {x^{\frac{3}{2}}} + {y^{\frac{3}{2}}} = {a^{\frac{3}{2}}} \cr
& \Rightarrow \frac{3}{2}.{x^{\frac{1}{2}}} + \frac{3}{2}.{y^{\frac{1}{2}}}\frac{{dy}}{{dx}} = 0\,\,\,\,\,\, \Rightarrow \frac{{dy}}{{dx}} = - \frac{{{x^{\frac{1}{2}}}}}{{{y^{\frac{1}{2}}}}} \cr
& \therefore {\left. {\,\frac{{dy}}{{dx}}} \right)_{\alpha ,\,\beta }} = 1\,\,\,\, \Rightarrow {\alpha ^{\frac{1}{2}}} + {\beta ^{\frac{1}{2}}} = 0 \cr
& {\text{Also }}{\alpha ^{\frac{3}{2}}} + {\beta ^{\frac{3}{2}}} = {a^{\frac{3}{2}}}\,\,\,\,\,\,\left\{ {\because \left( {\alpha ,\,\beta } \right)\,{\text{is on the curve}}} \right\} \cr
& {\text{These give no values of }}\alpha ,\,\beta \cr
& {\left. {\,{\text{Now, }}\frac{{dy}}{{dx}}} \right)_{\alpha ,\,\beta }} = - 1\,\,\,\,\,\,\,\,\,\, \Rightarrow \,{\alpha ^{\frac{1}{2}}} = {\beta ^{\frac{1}{2}}} \cr
& {\text{Also }}{\alpha ^{\frac{3}{2}}} + {\beta ^{\frac{3}{2}}} = {a^{\frac{3}{2}}}\,\,\therefore \,\alpha = \beta = \frac{a}{{{2^{\frac{2}{3}}}}} \cr
& \therefore \,\,{\text{there is only one point}}{\text{.}} \cr} $$