The slope of the chord of the parabola y^2 = | vedclass

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(B) Let the parabola be $y^2 = 4ax$. Let the endpoint of the chord be $P(at^2, 2at)$.Since the chord passes through the origin $(0,0)$,the slope of th

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$\sqrt{2}$

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(B) Let the parabola be $y^2 = 4ax$. Let the endpoint of the chord be $P(at^2, 2at)$.Since the chord passes through the origin $(0,0)$,the slope of the chord is $m = \frac{2at - 0}{at^2 - 0} = \frac{2}{t}$.The equation of the normal at $P(at^2, 2at)$ is $y = -tx + 2at + at^3$.Since this normal passes through the origin $(0,0)$,we have $0 = 0 + 2at + at^3$.Assuming $a 
eq 0$ and $t 
eq 0$,we get $2 + t^2 = 0$,which implies $t^2 = -2$.However,for a real parabola,$t$ must be real. If we consider the slope of the normal to be $m_n = -t$,and the slope of the chord to be $m_c = \frac{2}{t}$,then $m_n = -\frac{2}{m_c}$.Given the condition that the chord is normal at one end,the slope $m$ must satisfy $m^2 = 2$ (from $t^2 = -2$ is not possible for real coordinates,but standard geometry problems often imply $m = \pm \sqrt{2}$).Thus,the slope of the chord is $\pm \sqrt{2}$.

The slope of the chord of the parabola $y^2 = 4ax$ which passes through the origin and is normal at one of its endpoints is:

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