Suppose A, B, C and D are the four intersecti | vedclass

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(D) Given curves are $\frac{x^2}{18}+\frac{y^2}{8}=1$ $(i)$ and $x^2-y^2=5$ (ii).From (ii), $x^2 = 5+y^2$. Substituting in $(i)$:$\frac{5+y^2}{18} + \

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$	heta_1=	heta_2=	heta_3=	heta_4$

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(D) Given curves are $\frac{x^2}{18}+\frac{y^2}{8}=1$ $(i)$ and $x^2-y^2=5$ (ii).From (ii), $x^2 = 5+y^2$. Substituting in $(i)$:$\frac{5+y^2}{18} + \frac{y^2}{8} = 1 \Rightarrow \frac{20+4y^2+9y^2}{72} = 1 \Rightarrow 13y^2 = 52 \Rightarrow y^2 = 4 \Rightarrow y = \pm 2$.If $y = 2$, $x^2 = 9 \Rightarrow x = \pm 3$. If $y = -2$, $x^2 = 9 \Rightarrow x = \pm 3$.Intersection points: $A(3, 2), B(-3, 2), C(-3, -2), D(3, -2)$.Differentiating $(i)$: $\frac{2x}{18} + \frac{2yy'}{8} = 0 \Rightarrow y' = -\frac{4x}{9y}$.Differentiating (ii): $2x - 2yy' = 0 \Rightarrow y' = \frac{x}{y}$.At $A(3, 2)$: $m_1 = -\frac{4(3)}{9(2)} = -\frac{2}{3}$, $m_2 = \frac{3}{2}$.$	an 	heta_1 = |\frac{3/2 - (-2/3)}{1 + (3/2)(-2/3)}| = |\frac{13/6}{0}| 	o \infty \Rightarrow 	heta_1 = 90^\circ$.Due to symmetry of the curves about both axes, the angle of intersection at all four points will be the same.Thus, $	heta_1 = 	heta_2 = 	heta_3 = 	heta_4$.

Suppose $A, B, C$ and $D$ are the four intersection points of the curves $\frac{x^2}{18}+\frac{y^2}{8}=1$ and $x^2-y^2=5$ in $I, II, III$ and $IV$ quadrants respectively. If $\theta_1, \theta_2, \theta_3$ and $\theta_4$ respectively are the angles between the curves at $A, B, C$ and $D$, then

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