Question

The line $4y-3x+48=0$ cuts the curve $y^2=64x$ in $A$ and $B$. If $AB$ subtends an angle $\theta$ at the origin, then $\tan \theta =$

• A. $\frac{20}{9}$
• B. $\frac{10}{9}$
• C. $\frac{5}{9}$
• D. $\frac{40}{9}$

Answer 1

The correct option is A $\frac{20}{9}$

$4y-3x+48=0$
$\frac{3x-4y}{48}=1$
$\therefore y^2-64x\left(\frac{3x-4y}{48}\right)=0$ [By homogenisation.]
$y^2-\frac{4}{3}x(3x-4y)=0$
$y^2-4x^2+\frac{16}{3}yx=0$
$3y^2-12x^2+16xy=0$
$\therefore \tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$
$=\left|\frac{2\sqrt{64+36}}{(3-12)}\right|$
$=\left|\frac{20}{-9}\right|$
$\tan \theta =\frac{20}{9}$ is a angle subtended by $AB$ at origin.