question_answer

If the slope of the curve \[2{{y}^{2}}=a{{x}^{2}}+b\] at \[(1,\,\,-\,1)\] is \[-\,1,\] find the values of a and b.

Answer:

Given equation of curve is \[2{{y}^{2}}=a{{x}^{2}}+b\]            ?(i) On differentiating Eq. (i) w.r.t. x., we get \[4y\frac{dy}{dx}=2ax\Rightarrow \frac{dy}{dx}=\frac{2ax}{4y}\] \[\therefore \]Slope of tangent at point \[(1,\,\,-\,1)\] is\[-\,1\]. \[\therefore \]      \[\frac{dy}{dx}=-\,1\Rightarrow \frac{2ax}{4y}=-1\] At point \[(1,\,\,-\,1)\], \[\frac{2a\,(1)}{4\,(-\,1)}=-\,1\Rightarrow \frac{a}{-\,2}\Rightarrow -\,1\Rightarrow a=2\] On putting the value of a in Eq. (i) and point \[(1,\,\,-\,1)\], we get \[2\,{{(-\,1)}^{2}}=2\,{{(1)}^{2}}+b\Rightarrow 2=2+b\]\[\Rightarrow \]\[b=0\]