Question

The value of ${}^{\prime }{a}^{\prime }$ 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/are

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

Answer 1

Answer: (B), (D)

The correct options are
B $-1$
D $1$

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