A) Is a constant dependent on a
B) Is a constant dependent on b
C) Is a constant independent of a and b
D) 0
Correct Answer: C
Solution :
| [c] Given relation is \[{{(x-a)}^{2}}+{{(y-b)}^{2}}={{c}^{2}},c>0\] |
| Let \[x-a=c\,\,\cos \theta \] and \[y-b=c\sin \theta \]. Therefore, |
| \[\frac{dx}{d\theta }=-c\sin \theta \] and \[\frac{dy}{d\theta }=c\cos \theta \] |
| \[\therefore \frac{dy}{dx}=-\cot \theta \] |
| Differentiating both sides with respect to \[\theta \], we get |
| \[\frac{d}{d\theta }\left( \frac{dy}{dx} \right)=\frac{d}{d\theta }(-cot\theta )\] |
| or \[\frac{d}{dx}\left( \frac{dy}{dx} \right)\frac{dx}{d\theta }=\cos e{{c}^{2}}\theta \] |
| or \[\frac{{{d}^{2}}y}{d{{x}^{2}}}(-c\,sin\theta )=\cos e{{c}^{2}}\theta \] |
| or \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{\cos e{{c}^{2}}\theta }{c}\] |
| \[\therefore \frac{{{\left[ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right]}^{\frac{3}{2}}}}{\frac{{{d}^{2}}y}{d{{x}^{2}}}}=\frac{c{{\left[ 1+{{\cot }^{2}}\theta \right]}^{\frac{3}{2}}}}{-\cos e{{c}^{3}}\theta }=\frac{c{{(cose{{c}^{2}}\theta )}^{\frac{3}{2}}}}{-\cos e{{c}^{3}}\theta }\] |
| \[=-c.\] |
| Which is constant and is independent of a and b. |
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