An angle of intersection of the curves,frac{x | vedclass

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(A) Given curves are $C_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and $C_2: x^2 + y^2 = ab$. Let the point of intersection be $(x_1, y_1)$. For $C_1$,

Vedclass · Accepted Answer

$	an ^{-1}\left(\frac{a-b}{\sqrt{ab}}ight)$

Vedclass · Answer

(A) Given curves are $C_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and $C_2: x^2 + y^2 = ab$. Let the point of intersection be $(x_1, y_1)$. For $C_1$,differentiating with respect to $x$: $\frac{2x}{a^2} + \frac{2yy'}{b^2} = 0 \implies y'_1 = -\frac{b^2 x_1}{a^2 y_1}$. For $C_2$,differentiating with respect to $x$: $2x + 2yy' = 0 \implies y'_2 = -\frac{x_1}{y_1}$. Solving the equations for intersection: $x_1^2 = \frac{a^2 b}{a+b}$ and $y_1^2 = \frac{a b^2}{a+b}$. The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{y'_1 - y'_2}{1 + y'_1 y'_2} ight|$. Substituting the slopes: $	an 	heta = \left| \frac{-\frac{b^2 x_1}{a^2 y_1} + \frac{x_1}{y_1}}{1 + \frac{b^2 x_1^2}{a^2 y_1^2}} ight| = \left| \frac{x_1 y_1 (a^2 - b^2)}{a^2 y_1^2 + b^2 x_1^2} ight|$. Substituting $x_1^2$ and $y_1^2$: $	an 	heta = \left| \frac{a-b}{\sqrt{ab}} ight|$. Since $a > b$,$	heta = 	an^{-1} \left( \frac{a-b}{\sqrt{ab}} ight)$.

An angle of intersection of the curves,$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and $x^{2}+y^{2}=ab$,where $a > b$,is :

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