The chord of the curve y=x^{2}+2ax+b joining | vedclass

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(D) Given the curve $y=x^{2}+2ax+b$.At $x=\alpha$, the point is $P(\alpha, \alpha^{2}+2a\alpha+b)$.At $x=\beta$, the point is $Q(\beta, \beta^{2}+2a\b

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$\frac{\alpha+\beta}{2}$

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(D) Given the curve $y=x^{2}+2ax+b$.At $x=\alpha$, the point is $P(\alpha, \alpha^{2}+2a\alpha+b)$.At $x=\beta$, the point is $Q(\beta, \beta^{2}+2a\beta+b)$.The slope of the chord $PQ$ is given by $m_{chord} = \frac{(\beta^{2}+2a\beta+b) - (\alpha^{2}+2a\alpha+b)}{\beta-\alpha}$.$m_{chord} = \frac{(\beta^{2}-\alpha^{2}) + 2a(\beta-\alpha)}{\beta-\alpha} = \frac{(\beta-\alpha)(\beta+\alpha) + 2a(\beta-\alpha)}{\beta-\alpha} = \alpha+\beta+2a$.The slope of the tangent to the curve at any point $x$ is $m_{tangent} = \frac{dy}{dx} = 2x+2a$.Since the chord is parallel to the tangent, their slopes must be equal:$2x+2a = \alpha+\beta+2a$.Solving for $x$, we get $2x = \alpha+\beta$, which implies $x = \frac{\alpha+\beta}{2}$.

The chord of the curve $y=x^{2}+2ax+b$ joining the points where $x=\alpha$ and $x=\beta$ is parallel to the tangent to the curve at abscissa $x$ equal to:

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