Question

Let $f:R\to R$ be a bijection. A curve represented by $y=f(x)$ is such that $f^{\prime }(x)>0\forall x\in R$. The tangent and normal drawn at $P(\alpha ,1)$ on the curve cuts the $X$-axis at $A,B$ respectively and $C$ is the foot of the perpendicular from $P$ onto the $X$-axis. If $P(\alpha ,1)$ is such a point that $AC+CB$ is minimum, then the tangent at $P$ is parallel to the line

• A. $x-y=0$
• B. $\alpha x+y-1=0$
• C. $j$
• D. $\frac{2x}{\alpha }-y={\alpha }^2$

Answer 1

Answer: (A)

Given, $y=f(x)$ Equation of tangent of curve $y=f(x)$ at $P(\alpha ,1)$ is
$y-1=f(\alpha )(x-\alpha )$
Equation of normal of curve at $P(\alpha ,1)$ is
$y-1=-\frac{1}{f^{\prime }(\alpha )}(x-a)$
$\therefore$ Tangent cut of $X$-axis at $A$
$\therefore A=\left(\alpha -\frac{1}{f^{\prime }\alpha },0\right)$
$\therefore$ Normal cut of $X$-axis at $B$
$\therefore B=(\alpha +f^{\prime }(\alpha ),0$ Point $C=(\alpha ,0)$
$AC+BC=\alpha -\alpha +\frac{1}{f^{\prime }(\alpha )}+\alpha +f^{\prime }(\alpha )-\alpha$
$AC-BC=\frac{1}{f^{\prime }(\alpha )}+f^{\prime }(\alpha )$
$AC+BC$ is minimum
When, $f^{\prime }(\alpha )=1$
$\therefore$ Equation of tangent of curve
$y-1=1(x-\alpha )\Rightarrow x-y=\alpha +1$
$\therefore$ Equation of tangent of curve is parallel to
$x-y=0$