If θ is the acute angle between the curves x²+y²=4 and y²=3x then θ=

Correct Answer is : 53{5}{{3}}35

Correct Answer is : 53{5}{{3}}35

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TS EAMCET 19-July-2022 Shift 1 Solved Paper

Section: Mathematics

Question : 71 of 160

Marks: +1, -0

\theta

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

y^{2}=3x

\tan \theta=

Solution: 👈: Video Solution

Given, equation of curves,

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

and

y^{2}=3x

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

\Rightarrow\; x^{2}+4x-x-4\;=0.

\Rightarrow\; x(x+4)-1(x+4)\;=0

\Rightarrow\; (x+4)(x-1)\;=0

\Rightarrow\; x\;=1

[x \neq -4]

y = \pm \sqrt{3}

The intersecting points are

(1, \sqrt{3})

and

(1, -\sqrt{3})

\therefore \;\; x^{2}+y^{2}=4

On differentiating w.r.t.

x

, we get

2x+2y\; \frac{dy}{dx}=0

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

\frac{dy}{dx}

(1, \sqrt{3})=\frac{-1}{\sqrt{3}} (\text{ let } m_1)

\frac{dy}{dx}

(1, -\sqrt{3})=\frac{1}{\sqrt{3}} (\text{ let } m_2)

y^{2}=3x \Rightarrow \frac{dy}{dx}=\frac{3}{2y}

\frac{dy}{dx}

x=(1, \sqrt{3})=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2} (\text{ let } m_3)

\frac{dy}{dx}

x=(1, -\sqrt{3})=\frac{-3}{2\sqrt{3}}=\frac{-\sqrt{3}}{2} (\text{ let } m_4)

\;\tan \theta = \left| \frac{m_1-m_3}{1+m_1 m_3} \right| = \left| \frac{-\frac{1}{2\sqrt{3}} - \sqrt{3}}{1 - \frac{1}{2}} \right| = \left| \frac{-5}{\sqrt{3}} \right| = \frac{5}{\sqrt{3}}

Also,

\tan \theta = \left| \frac{m_2-m_4}{1+m_2 m_4} \right| = \left| \frac{\frac{1}{2\sqrt{3}} - \sqrt{3}}{1 - \frac{1}{2}} \right| = \frac{5}{\sqrt{3}}

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