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

Correct Answer is : 563{5}{6{3}}635

Correct Answer is : 563{5}{6{3}}635

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TS EAMCET 20-July-2022 Shift 2 Solved Paper

Section: Mathematics

Question : 71 of 160

Marks: +1, -0

\theta

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

y^{2}=3 x

\tan \theta =

Solution:

Curve I

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

On differentiating w.r.t.

x

, we get

\;2x-2y y^{'}=0

\Rightarrow \;\; x=y y^{'}

\Rightarrow \;\; y^{'} = \frac{x}{y} \ldots \text{(i)}

\Rightarrow \;\; m_1 = \frac{x}{y}

Curve II

y^{2}=3x

On differentiating w.r.t.

x

, we get

\;2y y^{'} = 3 \times 1

\Rightarrow \; y^{'} = \frac{3}{2y}

\Rightarrow \; m_2 = \frac{3}{2y}

.....(ii)

To get intersection point,

x^{2}-4=y^{2},\; y^{2}=3x

On putting

x^{2}-4=3x

\;x^{2}-3x-4=0

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

\; \Rightarrow \;\; x(x-4)+1(x-4)=0 \; \text{". "}\;

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

\;x=4,-1

From the graph it is clear that

x=4

\therefore \;\; y^{2}=3 \times 4

\Rightarrow \;\; y = \pm 2 \sqrt{3}

At point

(4, 2\sqrt{3})

from Eqs. (i) and (ii), we get

m_1 = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}}

m_2 = \frac{3}{2(2\sqrt{3})} = \frac{\sqrt{3}}{4}

\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|

\tan \theta = \left| \frac{ \frac{2}{\sqrt{3}} - \frac{\sqrt{3}}{4} }{ 1 + \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{4} } \right| = \left| \frac{ \frac{8-3}{4\sqrt{3}} }{ \frac{4\sqrt{3}+2\sqrt{3}}{4\sqrt{3}} } \right| = \frac{5}{6\sqrt{3}}

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