Question

lf the chord joining the points where $x=p,x=q$ on the curve $y=a{x}^2+bx+c$ is parallel to the tangent drawn to the curve at $(\alpha ,\beta )$ then $\alpha =$

• A. $2pq$
• B. $\sqrt{pq}$
• C. $\frac{p+q}{2}$
• D. $\frac{p-q}{2}$

Answer 1

The correct option is C $\frac{p+q}{2}$

$y^{\prime }=2ax+b$
$y^{\prime }{|}_{x=b}=(2a\alpha +b)$
$2a\alpha +b=\left(\frac{y_2-y_1}{x_2-x_1}\right)$
$2a\alpha +b=\frac{a{p}^2+bp-a{q}^2-bq}{(p-q)}$
$=\frac{a(p-q)(p+q)+b(p-a)}{(p-q)}$
$2a\alpha +b=a(p+q)+b$
$\alpha =\left(\frac{p+q}{2}\right)$