Question

The angle between the curves ${{x}^{2}}+{{y}^{2}}=25$ and ${{x}^{2}}+{{y}^{2}}-2x+3y-43=0$ at
(-3,4) is
(a) ${{\tan }^{-1}}\left( 1 \right)$
(b) ${{\tan }^{-1}}\left( \dfrac{1}{68} \right)$
(c) $\dfrac{\pi }{2}$
(d) ${{\tan }^{-1}}\left( \dfrac{3}{4} \right)$

Answer 1

Hint: In this question, we first need to find the slope of the curves by differentiating the curves at the given point and then substituting the respective coordinates of the given point.Then from the formula for angle between the curves in terms of slopes, we can get the result by substituting the respective values in the formula and simplifying.
Formula for the angle between two curves:
$\theta ={{\tan }^{-1}}\left( \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right)$

Complete step by step answer:
Let us assume the slopes of the given two curves as m1 and m2.
As we already know that the slope of a curve is given by
$m=\dfrac{dy}{dx}$
Now, on considering the first curve given in the question we get,
$\Rightarrow {{x}^{2}}+{{y}^{2}}=25$
Now, on differentiating it with respect to x on both sides we get,
$\Rightarrow 2x+2y\times \dfrac{dy}{dx}=0$
Now, on rearranging and cancelling the common terms we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{-x}{y}$
Now, slope of the curve at the given point (-3,4) is
$\Rightarrow {{m}_{1}}=\dfrac{-x}{y}$
Let us now substitute the respective values of x and y.

& \Rightarrow {{m}_{1}}=\dfrac{-\left( -3 \right)}{4} \\\ & \therefore {{m}_{1}}=\dfrac{3}{4} \\\ \end{aligned}$$ Now, on considering the second curve given in the question we get, $$\Rightarrow {{x}^{2}}+{{y}^{2}}-2x+3y-43=0$$ Now, on differentiating it with respect to x on both sides we get, $$\Rightarrow 2x+2y\times \dfrac{dy}{dx}-2+3\dfrac{dy}{dx}=0$$ Now, on writing the differential terms together in the above equation we get, $$\Rightarrow \left( 2y+3 \right)\times \dfrac{dy}{dx}+2x-2=0$$ Now, on rearranging the terms in the above equation we get, $$\Rightarrow \dfrac{dy}{dx}=\dfrac{2-2x}{2y+3}$$ Now, the slope of the curve at the given point (-3,4) is $$\Rightarrow {{m}_{2}}=\dfrac{2-2x}{2y+3}$$ Let us now substitute the respective values of x and y in the above equation. $$\Rightarrow {{m}_{2}}=\dfrac{2-2\times \left( -3 \right)}{2\times 4+3}$$ Now, on the further simplification we get, $$\begin{aligned} & \Rightarrow {{m}_{2}}=\dfrac{2+6}{8+3} \\\ & \therefore {{m}_{2}}=\dfrac{8}{11} \\\ \end{aligned}$$ As we already know the formula for the angle between two curves we get, $$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right)$$ Now, by substituting the respective values of slopes in the above equation we get, $$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\dfrac{3}{4}-\dfrac{8}{11}}{1+\dfrac{3}{4}\times \dfrac{8}{11}} \right)$$ Now, by taking the L.C.M. and on further simplification we get, $$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\dfrac{33-32}{44}}{\dfrac{44+24}{44}} \right)$$ $$\therefore \theta ={{\tan }^{-1}}\left( \dfrac{1}{68} \right)$$ Hence, the correct option is (b). Note: It is important to note that the slope of the curve at a given point can be found by differentiating it with respect to x and then substituting the respective coordinates of the point in the equation so obtained.While calculating the slopes and when differentiating we should not neglect any of the terms and should not write the wrong differentiation value because it changes the corresponding slope of the curve which in turn also changes the angle between the curves.