Question

The value of $a$ for which the area of the triangle included between the axes and any tangent to the curve $x^{a}y={\lambda }^a$ is constant, is:

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $-1$
• D. $1$

Answer 1

Answer: (C), (D)

$-1$
$1$

Given curve $x^{a}y={\lambda }^a$
$(\lambda ,1)$ is a point on the given curve.
Now, differentiating w.r.t. $x$ we get
$a{x}^{a-1}y+x^a\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{-a{x}^{a-1}y}{x^a}=\frac{-ay}{x}$
at $(\lambda ,1),\frac{dy}{dx}=-\frac{a}{\lambda }$
Equation of tangent at $(\lambda ,1)$
$y-1=\frac{-a}{\lambda }(x-\lambda )$
Now, $x=0,\Rightarrow y=1+a$
$y=0\Rightarrow x=\frac{\lambda }{a}+\frac{\lambda (1+a)}{a}$
Area$=A=\frac{1}{2}\times (1+a)\frac{\lambda (1+a)}{a}$
Now, $\frac{dA}{da}=\frac{1}{2}\lambda \left[\frac{a\times 2(1+a)-{(1+a)}^2}{a^2}\right]=0$
$\Rightarrow (2a-1-a)(1+a)=0$
$\Rightarrow (a-1)(a+1)=0$
$\therefore a=1,-1$