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Question: The acute angle between the curves \[y = \left| {{x^2} - 1} \right|\] and \[y = \left| {{x^2} - 3} \...

The acute angle between the curves y=x21y = \left| {{x^2} - 1} \right| and y=x23y = \left| {{x^2} - 3} \right| at their points of intersection when x>0x > 0 .

Explanation

Solution

Here, you can roughly draw the graph to get the acute angle between the curves and where both of the graphs intersects. They draw a tangent line to get the point of intersection and then find the slope and the equation of the straight line. Remember x>0x > 0 .

Formula used: For finding the slope we will find dydx=m\dfrac{{dy}}{{dx}} = m , for equation of straight line yy1=m(xx1)y - {y_1} = m\left( {x - {x_1}} \right) , acute angle between the curves tan(θθ1)=m1m21+m1m2\tan \left( {\theta - {\theta _1}} \right) = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|.

Complete step by step solution:
We will first draw the graphs for y=x21y = \left| {{x^2} - 1} \right| and y=x23y = \left| {{x^2} - 3} \right| for x>0x > 0 in a single graph. So, that both the graphs intersect each other and draw the tangents on the graph.

As, it is clear x>1x > 1 So, y=x21y = {x^2} - 1 and x<3x < \sqrt 3 so modulus of y=x23y = \left| {{x^2} - 3} \right| will open negative. As, y=(x23)y = - \left( {{x^2} - 3} \right) .
Comparing both the equations and calculating point of intersection.

\Rightarrow {x^2} - 1 = $$$$ - \left( {{x^2} - 3} \right)
Self Made

\Rightarrow {x^2} - 1 = $$$$ - {x^2} + 3
On Simplifying,
2x2=4\Rightarrow 2{x^2} = 4
x2=2\Rightarrow {x^2} = 2
We get x=2x = \sqrt 2
Putting x=2x = \sqrt 2 in y=x21y = {x^2} - 1
y=1\therefore y = 1
So, point of intersection are: (2,1)\left( {\sqrt 2 ,1} \right)
Firstly, we will calculate slope for y=x21y = {x^2} - 1
dydx=2x\dfrac{{dy}}{{dx}} = 2x
Put x=2x = \sqrt 2
We get,
dydx=22=m1\dfrac{{dy}}{{dx}} = 2\sqrt 2 = {m_1}
secondly, we will calculate slope for y=x2+3y = - {x^2} + 3
dydx=2x\dfrac{{dy}}{{dx}} = - 2x
Put x=2x = \sqrt 2
We get,
dydx=22=m2\dfrac{{dy}}{{dx}} = - 2\sqrt 2 = {m_2}
Now we will calculate equation of straight line using formula yy1=m(xx1)y - {y_1} = m\left( {x - {x_1}} \right)
For equation 1 m1=22{m_1} = 2\sqrt 2 and (x1,y1)=(2,1)\left( {{x_1},{y_1}} \right) = \left( {\sqrt 2 ,1} \right)
y1=22(x2)y - 1 = 2\sqrt 2 \left( {x - \sqrt 2 } \right)

For equation 2 m2=22{m_2} = - 2\sqrt 2 and (x1,y1)=(2,1)\left( {{x_1},{y_1}} \right) = \left( {\sqrt 2 ,1} \right)
y1=22(x2)y - 1 = - 2\sqrt 2 \left( {x - \sqrt 2 } \right)
Now, we will calculate the angle between the curves using the formula tan(θθ1)=m1m21+m1m2\tan \left( {\theta - {\theta _1}} \right) = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right| where m1=22{m_1} = 2\sqrt 2 and m2=22{m_2} = - 2\sqrt 2
tan(θθ1)=22(22)1+(22)(22)\tan \left( {\theta - {\theta _1}} \right) = \left| {\dfrac{{2\sqrt 2 - \left( { - 2\sqrt 2 } \right)}}{{1 + \left( {2\sqrt 2 } \right)\left( { - 2\sqrt 2 } \right)}}} \right|
On simplifying we get,
tan(θθ1)=22+221(22)(22)\tan \left( {\theta - {\theta _1}} \right) = \left| {\dfrac{{2\sqrt 2 + 2\sqrt 2 }}{{1 - \left( {2\sqrt 2 } \right)\left( {2\sqrt 2 } \right)}}} \right|
tan(θθ1)=4218\tan \left( {\theta - {\theta _1}} \right) = \left| {\dfrac{{4\sqrt 2 }}{{1 - 8}}} \right|
Hence, we get tan(θθ1)=427\tan \left( {\theta - {\theta _1}} \right) = \left| {\dfrac{{4\sqrt 2 }}{{ - 7}}} \right|
As, modulus is there so it will open as a positive sign.

tan(θθ1)=427\tan \left( {\theta - {\theta _1}} \right) = \dfrac{{4\sqrt 2 }}{7}
So, (θθ1)=tan1(427)\left( {\theta - {\theta _1}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{4\sqrt 2 }}{7}} \right) i.e (θθ1)\left( {\theta - {\theta _1}} \right) = angle between the curves

Note:
As in these types of questions we must remember that from a graph we can easily calculate the point of intersection and we have to draw two tangents and write the equation accordingly and the equations make the straight line and hence an acute angle can be calculated when x>0x > 0.