Question

The chord joining the points where x = p and x = q on the curve $y=a{x}^2+bx+c$ is parallel to the tangent at the point on the curve whose abscissa is

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

Answer 1

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

Given $f\left(x\right)=a{x}^2+bx+c$ is continuous on $[p,q]$ and

$f{}^{\prime }\left(x\right)=2ax+b$ is differentiable on $(p,q)$.

Therefore, Lagrange's mean value theorem can be applied.

Lagrange's mean value theorem states that if $f\left(x\right)$ be continuous on $[a,b]$ and differentiable on $(a,b)$ then there exists some $c$ between $a$ and $b$ such that $f{}^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Therefore, $f{}^{\prime }\left(c\right)=\frac{(a{q}^2+bq+c)-(a{p}^2+bp+c)}{q-p}$

$⟹2ac+b=\frac{a(q^2-p^2)+b(q-p)}{q-p}$

$⟹2ac+b=a(q+p)+b$

$⟹2ac=a(p+q)$

$⟹2c=p+q$

$⟹c=\frac{p+q}{2}$

Therefore, the abscissa of the point on the curve where the tangent is parallel to the chord is $\frac{p+q}{2}$