Question

A point on the curve $f\left(x\right)=\sqrt{x^2-4}$ defined in $[2,4]$ where the tangent is parallel to the chord joining two points on the curve

• A. $(\sqrt{2},\sqrt{6})$
• B. $(\sqrt{6},\sqrt{2})$
• C. $(2,6)$
• D. $(6,2)$

Answer 1

The correct option is B $(\sqrt{6},\sqrt{2})$

Given $f\left(x\right)=\sqrt{x^2-4}$ is continuous on $[2,4]$ and

$f^{{}^{\prime }}\left(x\right)=\frac{x}{\sqrt{x^2-4}}$ is differentiable on $(2,4)$.

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, $f^{{}^{\prime }}\left(c\right)=\frac{\sqrt{4^2-4}-\sqrt{2^2-4}}{4-2}$

$⟹\frac{c}{\sqrt{c^2-4}}=\frac{\sqrt{12}}{2}$

$⟹\frac{c}{\sqrt{c^2-4}}=\sqrt{3}$

$⟹c^2=3(c^2-4)$

$⟹c^2-6=0$

$⟹c=\sqrt{6}$

Therefore, $f\left(c\right)=\sqrt{6-4}=\sqrt{2}$

Therefore, $(c,f(c\left)\right)=(\sqrt{6},\sqrt{2})$ is a point on the curve where the tangent is parallel to the chord.