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Q.
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 $x =$
Given curve,
$ y=x^{2}+2 a x+b $
At $ x=\alpha, $
$y =\alpha^{2}+2 a \alpha+b $
and at $ x =\beta$,
$ y=\beta^{2}+2 a \beta + b $
$\therefore $ Slope of line joining $\left(\alpha, \alpha^{2}+2 a \alpha+b\right)$ and $\left(\beta, \beta^{2}+2 a \beta+b\right)$ is
$=\frac{\left(\alpha^{2}+2 a \alpha+b\right)-\left(\beta^{2}+2 a \beta+b\right)}{\alpha-\beta}$
$=\frac{\alpha^{2}+2 a \alpha+b-\beta^{2}-2 a \beta-b}{\alpha-\beta} $
$=\frac{\left(\alpha^{2}-\beta^{2}\right)+2 a(\alpha-\beta)}{\alpha-\beta} $
$=\frac{(\alpha-\beta)(\alpha+\beta)+2 a(\alpha-\beta)}{\alpha-\beta} $
$=\alpha+\beta+2 a$
Slope of given curve $=\frac{d y}{d x}$
$=2 x+2 a$
Now, according to question, tangent is parallel to the chord. Therefore,
$2 x+2 a =\alpha+\beta+2 a $
$\Rightarrow 2 x =\alpha+\beta $
$\Rightarrow x=\frac{\alpha+\beta}{2}$