Question

If the tangent at the P of the curve $y^2=x^3$ intersect the curve again at Q and the straight lines OP, OQ make angles $\alpha ,\beta$ with the x-axis, where O is the origin. then, $\tan \alpha /\tan \beta$ has the value equal to

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

Answer 1

The correct option is B $-2$

Let the parametric coordinates at $P$ and $Q$ be :

$(t_1^2,t_1^3),(t_2^2,t_2^3)$

Then slope of the tangent at the point $P\Rightarrow \frac{dy}{dx}=\frac{3x^2}{2y}=\frac{3}{2t_1}$

Also, line joining $P$ and $Q$ would have a slope given by: $\frac{t_2^3-t_1^3}{t_2^2-t_1^2}$.

Equating both slopes, $2\frac{t_2^2}{t_1^2}=\frac{t_2}{t_1}+1$------(1)

Also. $\tan \alpha =t_1,\tan \beta =t_2$-------(2)

$\therefore$ From (1),

$2\frac{{\tan }^2\beta }{{\tan }^2\alpha }=\frac{\tan \beta }{\tan \alpha }+1$.

Let $\frac{t_2}{t_1}=x$

$\therefore 2x^2=x+1$

or, $2x^2-x-1=0$

or, $2x^2-2x+x-1=0$

or, $2x(x-1)+1(x-1)=0\Rightarrow (x-1)(2x+1)=0$

$\therefore x=1,1/2$

i.e $\frac{t_2}{t_1}=1\Rightarrow t_2=t_1$ or, $\frac{t_2}{t_1}=\frac{-1}{2}\Rightarrow t_1=-2t_2$

$\therefore$ From (2) if $t_2=t_1$; and if $t_1=2t_2$

$\tan \alpha =\tan \beta$ $\tan \alpha =-2\tan \beta$

$\therefore \frac{\tan \alpha }{\tan \beta }=1$ $\therefore \frac{\tan \alpha }{\tan \beta }=-2$