Question

If the tangent at P of the curve $y^2=x^3$ intersects the curve again at Q and the straight lines OP, OQ ma 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 $P$ be $(x_1,y_1)$ and $Q$ be $(x_2,y_2)$

equation of tangent $(x_2,y_2)$, so $\frac{y_0-y_1}{x_2-x_1}=\frac{3x_1^2}{2y_1}$

need to find out $\frac{\tan \alpha }{\tan \beta }=\frac{\frac{y_1}{x_1}}{y_2/x_2}=\frac{y_1{x}_2}{x_1{y}_2}$

$\frac{y_2-y_1}{x_2-x_1}=\frac{3x_1^2}{2y_1}$

$2y_1{y}_2-2y^2=3x_1^2{x}_2-3x_1^3$

$2y_1{y}_2-2y_1^2=3x_1^2{x}_2-3y^2$

$2y_1{y}_2+y_1^2=3x_1^2{x}_2$

$y_1^3(2y_2+y_1{)}^3=27x_1^6{x}_2^3$

$y_1^3(2y_2+y_1{)}^3=27y_1^4{y}_2^3$

$(2y_2+y_1{)}^3-27y_1{y}_2$

$8y_2^3-15y_2^2{y}_1+6y_2{y}_1^2+y_1^3=0$

$(y_2-y_1{)}^2(8y_2+y_1)=0$

$8y_2=-y_1\Rightarrow 64y_2^2=y_1^2$

$64x_2^3=x_1^3\Rightarrow 4x_2=x_1$

$\frac{\tan \alpha }{\tan \beta }=\frac{y_1{x}_2}{x_1{y}_2}$

$=\frac{-8y_2{x}_2}{4x_2{y}_2}=-2$