The angle between y^{2}=4x and x^{2}+y^{2}=12 | vedclass

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Question

(C) Given equations of the curves are:$y^{2}=4x$  $(i)$$x^{2}+y^{2}=12$  (ii)First,find the intersection points by substituting $(i)$ into (ii):$x^{2}

Vedclass · Accepted Answer

$	an^{-1} 2\sqrt{2}$

Vedclass · Answer

(C) Given equations of the curves are:$y^{2}=4x$  $(i)$$x^{2}+y^{2}=12$  (ii)First,find the intersection points by substituting $(i)$ into (ii):$x^{2}+4x=12 \Rightarrow x^{2}+4x-12=0$$(x+6)(x-2)=0$Since $y^{2}=4x$,$x$ must be $\ge 0$,so $x=2$.For $x=2$,$y^{2}=8 \Rightarrow y=\pm 2\sqrt{2}$.Intersection points are $(2, 2\sqrt{2})$ and $(2, -2\sqrt{2})$.Now,find the slopes $m_{1}$ and $m_{2}$ by differentiating:For $(i)$: $2y \frac{dy}{dx} = 4 \Rightarrow m_{1} = \frac{2}{y}$.For (ii): $2x + 2y \frac{dy}{dx} = 0 \Rightarrow m_{2} = -\frac{x}{y}$.At $(2, 2\sqrt{2})$:$m_{1} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$ and $m_{2} = -\frac{2}{2\sqrt{2}} = -\frac{1}{\sqrt{2}}$.The angle $	heta$ is given by $	an 	heta = \left| \frac{m_{1}-m_{2}}{1+m_{1}m_{2}} ight|$.$	an 	heta = \left| \frac{\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}})}{1 + (\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})} ight| = \left| \frac{\frac{2}{\sqrt{2}}}{1 - \frac{1}{2}} ight| = \left| \frac{\sqrt{2}}{\frac{1}{2}} ight| = 2\sqrt{2}$.Thus,$	heta = 	an^{-1}(2\sqrt{2})$.

Column-$I$	Column-$II$
$A$. Circle	$P$. Locus of point $(h, k)$ such that the line $hx + ky = 1$ touches the circle $x^2 + y^2 = 4$
$B$. Parabola	$Q$. Point $z$ in the complex plane satisfies $\|z + 2\| - \|z - 2\| = \pm 3$
$C$. Hyperbola	$R$. Eccentricity of the conic lies in the interval $[1, \infty)$
	$S$. Point $z$ in the complex plane satisfies $Re(z + 1)^2 = \|z\|^2 + 1$

The angle between $y^{2}=4x$ and $x^{2}+y^{2}=12$ at a point of their intersection is

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