A) 1
B) \[\sqrt{3}\]
C) 3
D) \[\frac{1}{\sqrt{3}}\]
Correct Answer: C
Solution :
Given equations are \[{{x}^{2}}+{{y}^{2}}=5\] ... (i) and \[{{y}^{2}}=4x\] .... (ii) On solving equation (i) and (ii) we get \[x=5\,,1\] at \[x=-5,\,\,{{y}^{2}}=-20\] (imaginary value) \[\therefore \] at \[x=1,\text{ }{{y}^{2}}=4\text{ }\Rightarrow \text{ }y=\pm 2\] Hence point of intersection are (1,2) and (1,-2). On differentiating equation (i) with respect to x we get \[2x+2y\frac{dy}{dx}=0\] \[\Rightarrow \] \[\frac{dy}{dx}=-\frac{x}{y}\] \[\therefore \,\,{{m}_{1}}={{\left( \frac{dy}{dx} \right)}_{(1,2)}}=-\frac{1}{2}\] And on diffemtiating equation (ii) with respect to \[x\], we get \[2y\frac{dy}{dx}=4\,\,\,\Rightarrow \,\,\,\frac{dy}{dx}=\frac{2}{y},\,\,{{m}_{2}}={{\left( \frac{dy}{dx} \right)}_{(1,2)}}=\frac{2}{2}=1\] Now \[\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\] \[=\left| \frac{-\frac{1}{2}-1}{1-\frac{1}{2}} \right|=\left| \frac{-\frac{3}{2}}{\frac{1}{2}} \right|\]
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