A) \[\frac{4}{9}\]
B) \[\frac{7}{17}\]
C) \[\frac{8}{17}\]
D) \[\frac{8}{15}\]
Correct Answer: D
Solution :
\[y=10-{{x}^{2}}~~~y=2+{{x}^{2}}\] \[10\text{ }-{{x}^{2}}=\text{ }2\text{ }+{{x}^{2}}\] \[8\,\,=\,\,2{{x}^{2}}\] \[{{x}^{2}}=\text{ }4\] \[x\,\,=\,\,2,\,\,-2\] points are \[\left( 2,\text{ }6 \right),\text{ }\left( -2,\text{ }6 \right)\] \[{{\left( \frac{dy}{dx} \right)}_{(2,\,\,\,6)}}\,\,\,=\,\,2x\,\,=\,\,4\] \[{{\left( \frac{dy}{dx} \right)}_{(-\,2,\,\,\,6)}}\,\,\,=\,\,-2x\,\,=\,-\,4\] \[\tan \,\theta \,\,=\,\,\left| \frac{4-(-4)}{1+4(-\,4)} \right|\,\,=\,\,\left| \frac{8}{1-16} \right|\,\,=\,\,\frac{8}{15}\]
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