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(B) The given curve is $y=x^2$. When it is shifted by '$a$' units along the positive $X$-axis, the new curve becomes $y=(x-a)^2$.To find the intersect

Vedclass · Accepted Answer

$	an 	heta=\frac{2|a|}{\left|1-a^2ight|}$

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(B) The given curve is $y=x^2$. When it is shifted by '$a$' units along the positive $X$-axis, the new curve becomes $y=(x-a)^2$.To find the intersection point, we set the equations equal:$x^2 = (x-a)^2$$x^2 = x^2 - 2ax + a^2$$2ax = a^2$Since $a 
eq 0$, we get $x = \frac{a}{2}$.Substituting $x = \frac{a}{2}$ into $y=x^2$, we get $y = \frac{a^2}{4}$.The intersection point is $(\frac{a}{2}, \frac{a^2}{4})$.Now, find the slopes of the tangents at this point:For $y=x^2$, $\frac{dy}{dx} = 2x$. At $x=\frac{a}{2}$, $m_1 = 2(\frac{a}{2}) = a$.For $y=(x-a)^2$, $\frac{dy}{dx} = 2(x-a)$. At $x=\frac{a}{2}$, $m_2 = 2(\frac{a}{2}-a) = -a$.The angle $	heta$ between the curves is given by:$	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight| = \left| \frac{a - (-a)}{1 + (a)(-a)} ight| = \left| \frac{2a}{1 - a^2} ight| = \frac{2|a|}{|1 - a^2|}$.Thus, option $(b)$ is correct.

$y=x^2$ is the given curve. Imagine that this curve is dragged along the positive $X$-axis to a distance of '$a$' units. If the acute angle between the curves at two positions is $\theta$, then

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