An angle between the curves x²=3y and x²+y²=4 is

Correct Answer is : tan⁡−153^{-1}{5}{{3}}tan−135

Correct Answer is : tan⁡−153^{-1}{5}{{3}}tan−135

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AP EAMCET Engineering 26 Apr 2017 Shift 1 Paper

Section: Mathematics

Question : 68 of 160

Marks: +1, -0

x^2=3y

x^2+y^2=4

Solution:

Consider the given equation of curves.

x^{2}=3 y

And,

x^{2}+y^{2}=4

Substitute

3 y

for

x^{2}

in the above equation.

y^{2}+3 y-4=0

y(y+4)-1(y+4)=0

(y+4)(y-1)=0

y=-4,1

y=-4

x^{2}=-12

This is not possible.

x=\pm\sqrt{3}

The point of intersection are

(\sqrt{3}, 1)

and

(-\sqrt{3}, 1)

So,1

\frac{dy}{dx}= \frac{2x}{3}

m_{1}= \frac{2}{3}(\sqrt{3})

Also,1

\frac{dy}{dx} = -\frac{x}{y}

m_{2}= -\frac{x}{y}

=\frac{\sqrt{3}}{1}

=\sqrt{3}

The angle between the curve is calculated as,

\tan \theta = \left| \frac{m_{2}-m_{1}}{1+m_{1} m_{2}} \right|

=\left| \frac{-\sqrt{3} - \frac{2\sqrt{3}}{3}}{1+ \left(\frac{2\sqrt{3}}{3}\right)(-\sqrt{3})} \right|

=\frac{5}{\sqrt{3}}

\theta = \tan^{-1}\left(\frac{5}{\sqrt{3}}\right)

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