An angle of intersection of the curves, {x²}{a²}+{y²}{b²}=1 and x²+y²=a b, a > b, is [31 Aug 2021 Shift 2]

Correct Answer is : tan⁡−1(a−bab)^{-1}≤ft({a-b}{{ab}})tan−1(aba−b)

Correct Answer is : tan⁡−1(a−bab)^{-1}≤ft({a-b}{{ab}})tan−1(aba−b)

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Section: Mathematics

Question : 67 of 117

Marks: +1, -0

\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1

x^{2}+y^{2}=a b, a > b

Solution:

Given curves

\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1

...(i)

and

x^{2}+y^{2}=a b

...(ii)

From Eqs. (ii)

y^{2}=a b-x^{2}

From Eq. (i),

b^{2} x^{2}+a^{2}(a b-x^{2})=a^{2} b^{2}

(b^{2}-a^{2}) x^{2}=a^{2} b(b-a)

\Rightarrow x^{2}=\frac{a^{2} b}{a+b}

y^{2}=a b-\frac{a^{2} b}{a+b}=\frac{a b^{2}}{a+b}

Point of intersection

(\sqrt{\frac{a^{2} b}{a+b}}, \sqrt{\frac{a b^{2}}{a+b}})

Now, differentiating Eq. (i) w.r.t. x, we have

\frac{dy}{dx}=-\frac{b^{2} x}{a^{2} y}=m_{1}

(Let)

and differentiating Eq. (ii) w.r.t. x,

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

(Let)

Let angle be θ, Then,

\tan\theta = \left| \frac{m_{1}-m_{2}}{1+m_{1} m_{2}} \right| = \left| \frac{-\frac{b^{2} x}{a^{2} y} + \frac{x}{y}}{1+\frac{b^{2} x^{2}}{a^{2} y^{2}}} \right|

= \left| \frac{x y (a^{2}-b^{2})}{a^{2} b^{2}} \right| = \left| \sqrt{\frac{a^{3} b^{3}}{(a+b)^{2}}} \cdot \frac{a^{2}-b^{2}}{a^{2} b^{2}} \right|

= \left| \frac{a-b}{\sqrt{ab}} \right|

\Rightarrow \theta = \tan^{-1}\left(\frac{a-b}{\sqrt{ab}}\right)

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