If ax + by = 1 is a normal to the parabola y^ | vedclass

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(C) The equation of a normal to the parabola $y^2 = 4px$ with slope $m$ is given by $y = mx - 2pm - pm^3$.Given the normal equation $ax + by = 1$, we

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$pa^3 = b^2 - 2pab^2$

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(C) The equation of a normal to the parabola $y^2 = 4px$ with slope $m$ is given by $y = mx - 2pm - pm^3$.Given the normal equation $ax + by = 1$, we can rewrite it as $y = -\frac{a}{b}x + \frac{1}{b}$.Comparing the two equations, we get $m = -\frac{a}{b}$ and the constant term $-2pm - pm^3 = \frac{1}{b}$.Substituting $m = -\frac{a}{b}$ into the constant term equation:$-2p(-\frac{a}{b}) - p(-\frac{a}{b})^3 = \frac{1}{b}$$\frac{2pa}{b} + \frac{pa^3}{b^3} = \frac{1}{b}$Multiplying both sides by $b^3$, we get $2pab^2 + pa^3 = b^2$, which can be rearranged as $pa^3 = b^2 - 2pab^2$.

If $ax + by = 1$ is a normal to the parabola $y^2 = 4px$, then the condition is:

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