Question

Find the angle between two curves x^2+y^2=8 and x^2= 2y

Answer 1

Answer:

Step-by-step explanation:

$\tt{Given,\,\,\,curves\,\,\,are\,\,\,x^2+y^2=8\,and\,\,\,x^2=2y}$

Differentiating both the curves w.r.t x, to get their slopes

$\sf{2x+2y\dfrac{dy}{dx}=0\,and\,\,\,2x=2\dfrac{dy}{dx}}$

$\sf{\implies\,\dfrac{dy}{dx}=-\dfrac{x}{y}\,and\,\,\,\dfrac{dy}{dx}=x}$

$\sf{\implies\,\dfrac{dy}{dx}=m_{1}=-\dfrac{x}{y}\,and\,\,\,\dfrac{dy}{dx}=m_{2}=x}$

To know the point of intersection, solve the two curves

$\sf{y^2+2y=8}$

$\sf{\implies\,y^2+2y-8=0}$

$\sf{\implies\,y^2+4y-2y-8=0}$

$\sf{\implies\,y(y+4)-2(y+4)=0}$

$\sf{\implies\,(y-2)(y+4)=0}$

$\sf{\implies\,y=2\,\,\,\,or\,\,\,\,y=-4}$

$\sf{If\,\,\,y=-4\,\,\,,then,\,\,\,x^2=-8,\,\,\,which\,\,\,is\,\,\,not\,\,\,possible}$

So,

$\sf{y=2\implies\,x=\pm2}$

So, coordinates of their point of intersection are (2, 2) and (- 2, 2)

To obtain the angle between, let us take the point (2, 2)

So,

$\sf{m_{1}=-1\,\,\,\,\,and\,\,\,\,\,m_{2}=2}$

Now,

$\sf{tan(\theta)=\left|\dfrac{m_{2}-m_{1}}{1+m_{2}\,m_{1}}\right|}$

$\sf{\implies\,tan(\theta)=\left|\dfrac{2+1}{1-2}\right|}$

$\sf{\implies\,tan(\theta)=\left|\dfrac{3}{-1}\right|}$

$\sf{\implies\,tan(\theta)=3}$

$\sf{\implies\,\theta=tan^{-1}(3)}$