A) Cuts at right angle
B) Touch each other
C) Cut at an angle \[\frac{\pi }{3}\]
D) Cut at an angle \[\frac{\pi }{4}\]
Correct Answer: A
Solution :
| [a] Two curves cuts at right angle if product of their slopes is -1. |
| Two gives curves are |
| \[{{x}^{3}}-3x{{y}^{2}}+2=0\] ? (i) |
| and \[3{{x}^{2}}y-{{y}^{3}}-2=0\] ? (ii) |
| Differentiate equ. (i), |
| \[3{{x}^{2}}-3\left[ {{y}^{2}}+2xy\frac{dy}{dx} \right]=0\] |
| \[\Rightarrow 3({{x}^{2}}-{{y}^{2}})=6xy\frac{dy}{dx}\] |
| \[\Rightarrow {{m}_{1}}=\frac{dy}{dx}=\frac{3({{x}^{2}}-{{y}^{2}})}{6xy}\] |
| Differentiate eq. (ii), |
| \[3{{x}^{2}}y-{{y}^{3}}-2=0\] |
| \[\Rightarrow 3\left[ {{x}^{2}}\frac{dy}{dx}+2xy \right]-3{{y}^{2}}\frac{dy}{dx}=0\] |
| \[\Rightarrow {{x}^{2}}\frac{dy}{dx}+2xy-{{y}^{2}}\frac{dy}{dx}=0\] |
| \[\Rightarrow ({{x}^{2}}-{{y}^{2}})\frac{dy}{dx}=-2xy\] |
| \[\Rightarrow {{m}_{2}}=\frac{dy}{dx}=\frac{-2xy}{({{x}^{2}}-{{y}^{2}})}\] |
| \[\therefore {{m}_{1}}\times {{m}_{2}}=\frac{({{x}^{2}}-{{y}^{2}})}{2xy}\times \frac{-2xy}{({{x}^{2}}-{{y}^{2}})}\] |
| \[\Rightarrow {{m}_{1}}\times {{m}_{2}}=-1\] |
| i.e., curves cuts at right angle. |
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