Question

The chord joining the points $x=a$ and $x=b$ on the curve $y=p{x}^2+qx+r,p\ne 0$ is perpendicular to the normal at the point on the curve whose abscissa is

• A. $a-b$
• B. $\frac{a-b}{2}$
• C. $\frac{a+b}{2}$
• D. $a+b$

Answer 1

The correct option is C $\frac{a+b}{2}$

Chord is perpendicular to the normal.
$\Rightarrow$ Chord is parallel to the tangent.
Let $(c,y(c\left)\right)$ be the point on the curve where the tangent is parallel to the given chord.

$\Rightarrow \frac{y\left(b\right)-y\left(a\right)}{b-a}=2pc+q\Rightarrow \frac{p(b^2-a^2)+q(b-a)}{b-a}=2pc+q\Rightarrow p(b+a)+q=2pc+q\Rightarrow c=\frac{a+b}{2}$