Question

For the curve $b{y}^2=(x+a{)}^3$ the square of subtangent is proportional to
[RPET 1999]

• A. $ta{n}^{-1}\left(3\right)$
• B. $ta{n}^{-1}\left(2\right)$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is C $\frac{\pi }{2}$

Let $y_1=co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2ta{n}^{-1}x,$
$co{t}^{-1}\left(\frac{1-3x^2}{3x-x^3}\right)=3ta{n}^{-1}x\Rightarrow \frac{d{y}_1}{d{y}_2}=\frac{\left(\frac{d{y}_1}{dx}\right)}{\left(\frac{d{y}_2}{dx}\right)}=\frac{\left(\frac{2}{1+x^2}\right)}{\left(\frac{3}{1+x^2}\right)}=\frac{2}{3}$