question_answer

Find the angle of intersection of the curves \[{{y}^{2}}=4ax\] and \[{{x}^{2}}=4by.\]

Answer:

Given equation of curves are \[{{y}^{2}}=4ax\]                             ?(i) and       \[{{x}^{2}}=4by\]                             ...(ii) On solving Eqs. (i) and (ii), we get             \[{{\left( \frac{{{x}^{2}}}{4b} \right)}^{2}}=4ax\] \[\Rightarrow \] \[{{x}^{4}}=64\,\,a{{b}^{2}}x\] \[\Rightarrow \]   \[{{x}^{4}}-64a{{b}^{2}}x=0\] \[\Rightarrow \]   \[x({{x}^{3}}-64a{{b}^{2}})=0\] \[\Rightarrow \]   \[x=0\,\,\,\text{or}\,\,\,{{x}^{3}}=64a{{b}^{2}}\] \[\Rightarrow \]   \[x=0\,\,\,\text{or}\,\,\,x=4{{a}^{1/3}}{{b}^{2/3}}\] \[\Rightarrow \]   \[y=0\,\,\,\text{or}\,\,\,y=4{{a}^{2/3}}{{b}^{1/3}}\]   [using Eq. (i)] Thus, the point of intersections are (0, 0) and \[(4{{a}^{1/3}}{{b}^{2/3}},\,\,4{{a}^{2/3}}{{b}^{1/3}}).\] Now, consider \[{{y}^{2}}=4ax\]\[\Rightarrow \] \[2y\frac{dy}{dx}=4a\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{2a}{y}\] and       \[{{x}^{2}}=4by\] \[\Rightarrow \] \[2x=4b\frac{dy}{dx}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{x}{2b}\] So at (0, 0), the tangent to the curve \[{{y}^{2}}=4ax\] is parallel to Y-axis and tangent to the curve \[{{x}^{2}}=4by\] is parallel to X-axis. Slope of tangent to the curve (i)    \[{{m}_{1}}=\frac{2a}{4{{a}^{2/3}}{{b}^{1/3}}}=\frac{1}{2}{{\left( \frac{a}{b} \right)}^{1/3}}\] and Slope of tangent to the curve (ii),             \[{{m}_{2}}=\frac{4{{a}^{1/3}}{{b}^{2/3}}}{2b}=2{{\left( \frac{a}{b} \right)}^{1/3}}\] Hence, \[\tan \theta =\left| \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right|\]             \[=\left| \frac{2{{\left( \frac{a}{b} \right)}^{1/3}}-\frac{1}{2}{{\left( \frac{a}{b} \right)}^{1/3}}}{1+2{{\left( \frac{a}{b} \right)}^{1/3}}\frac{1}{2}{{\left( \frac{a}{b} \right)}^{1/3}}} \right|\]             \[=\left| \frac{{{\left( \frac{a}{b} \right)}^{1/3}}\left( 2-\frac{1}{2} \right)}{1+{{\left( \frac{a}{b} \right)}^{2/3}}} \right|=\left| \frac{\frac{3}{2}\cdot {{\left( \frac{a}{b} \right)}^{1/3}}}{\frac{{{b}^{2/3}}+{{a}^{2/3}}}{{{b}^{2/3}}}} \right|\]             \[=\left| \frac{\frac{3}{2}\times {{\left( \frac{a}{b} \right)}^{1/3}}\times {{b}^{2/3}}}{{{b}^{2/3}}+{{a}^{2/3}}} \right|=\left| \frac{3{{a}^{1/3}}\cdot {{b}^{1/3}}}{2({{b}^{2/3}}+{{a}^{2/3}})} \right|\]