Question

If the curves ${(x-20)}^2+{(y-20)}^2=r^2$ and $\frac{x^2}{4}+\frac{y^2}{1}=100$ intersects orthogonally at a point $(20\cos \theta ,10\sin \theta )$ then

• A. $\tan \frac{\theta }{2}=2$
• B. $\tan \frac{\theta }{2}=\sqrt{2}$
• C. $\tan \frac{\theta }{2}=\frac{1}{20}$
• D. $\tan \frac{\theta }{2}=10$

Answer 1

The correct option is A $\tan \frac{\theta }{2}=2$

$(X-20{)}^2+(Y-20{)}^2=R^2$

$2(X-20)+2(Y-20)M_1=0$

$M_1=-\frac{(x-20)}{y-20}=\frac{-2(1-\cos \theta )}{(\sin \theta -2)}$

$\frac{x^2}{4}+\frac{y^2}{1}=100$

$\frac{2x}{4}+2y{m}_2=0$

$m_2=\frac{-x}{4y}=\frac{-\cos theta}{2\sin \theta }$

here $m_1{m}_2=-1$

$\frac{2(1-\cos \theta )}{(\sin \theta -2)}\frac{\left(\cos \theta \right)}{\left(2\sin \theta \right)}=1$

$(2-2\cos \theta )\left(\cos \theta \right)=2\sin \theta (\sin \theta -2)$

solving

$\tan \theta =-4/3$

$\tan \left(\theta /2\right)=2$