A) \[\frac{1}{2}\]
B) \[\frac{1}{3}\]
C) 2
D) 3
Correct Answer: B
Solution :
We have the curves \[y=1-a{{x}^{2}}\] \[\Rightarrow \] \[\frac{dy}{dx}=-2ax\] \[\Rightarrow \] \[{{m}_{1}}=-2a{{x}_{1}}\]at \[({{x}_{1}},{{y}_{1}})\] and \[y={{x}^{2}}\] \[\Rightarrow \] \[\frac{dy}{dx}=2x\] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},{{y}_{1}})}}=2{{x}_{1}}\] \[\Rightarrow \] \[{{m}_{2}}=2{{x}_{1}}\] Since, both the curves intersect orthogonally. \[\therefore \] \[{{m}_{1}}{{m}_{2}}=-1\] \[\Rightarrow \] \[-2a{{x}_{1}}.2{{x}_{1}}=-1\] \[\Rightarrow \]\[4ax_{1}^{2}=1\] ?(i) Now, solving the given equations \[y=1-a{{x}^{2}}\]and \[y={{x}^{2}},\]we get the intersection point \[({{x}_{1}},{{y}_{1}})=\left( \frac{1}{\sqrt{1+a}},\frac{1}{(1+a)} \right)\] So, from Eq. (i), \[4a.\frac{1}{1+a}=1\] \[\Rightarrow \] \[4a=1+a\] \[\Rightarrow \] \[3a=1\] \[\Rightarrow \] \[a=\frac{1}{3}\]
You need to login to perform this action.
You will be redirected in
3 sec
