Question

Let $f: R \rightarrow R$ be a bijection. A curve represented by $y=f(x)$ is such that $f '(x) > 0 \forall x \in R$. 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 $A C+C B$ 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. $\frac{2 x}{\alpha}-y=\alpha^{2}$

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 $A$
$\therefore A=\left(\alpha-\frac{1}{f' \alpha}, 0\right)$
$\therefore $ Normal cut of $X$-axis at $B$
$\therefore B=\left(\alpha +f'(\alpha), 0\right.$ Point $C=(\alpha, 0)$
$A C+B C=\alpha-\alpha+\frac{1}{f^{\prime}(\alpha)}+\alpha +f'(\alpha)-\alpha$
$A C-B C=\frac{1}{f'(\alpha)}+f'(\alpha)$
$A C+B C$ is minimum
When, $f'(\alpha)=1$
$\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
$x-y=0$