Question

Let $Q(t_1^2,2t_1)$ and $(t_2^2,2t_2),$ then ordered pair $(t_1+t_2,t_1.t_2)$ is

• A. $(\frac{2t[(t^2+2){\tan }^2\alpha +1]}{1-t^2{\tan }^2\alpha },\frac{t^2-(t^2+2{)}^2{\tan }^2\alpha }{1-t^2{\tan }^2\alpha })$
• B. $(\frac{2t(t^2-2){\tan }^2\alpha -1}{1-t^2{\tan }^2\alpha },\frac{t^2-(t^2+2{)}^2{\tan }^2\alpha }{1-t^2{\tan }^2\alpha })$
• C. $(0,0)$
• D. $(\frac{2t(t^2+2){\tan }^2\alpha +1}{1-t^2{\tan }^2\alpha },\frac{t^2+(t^2+2{)}^2{\tan }^2\alpha }{1-t^2{\tan }^2\alpha })$

Answer 1

The correct option is A $(\frac{2t[(t^2+2){\tan }^2\alpha +1]}{1-t^2{\tan }^2\alpha },\frac{t^2-(t^2+2{)}^2{\tan }^2\alpha }{1-t^2{\tan }^2\alpha })$

Let any variable point on parabola be $({\lambda }^2,2\lambda )$
So, $\left|\begin{array}{c}\frac{\frac{2}{\lambda +t}-\frac{1}{t}}{1+\frac{2}{t(\lambda +t)}}\end{array}\right|=\tan \alpha$
$\Rightarrow {\lambda }^2(1-t^2{\tan }^2\alpha )-2\lambda t(1+(t^2+2\left){\tan }^2\alpha \right)+t^2-(t^2+2{)}^2{\tan }^2\alpha =0$
This equation in $\lambda$ has two roots $t^1\&t_2$
So, $t_1+t_2=\frac{2t(1+(t^2+2\left){\tan }^2\alpha \right)}{1-t^2{\tan }^2\alpha }$
and $t_1{t}_2=\frac{t^2-(t^2-2{)}^2{\tan }^2\alpha }{1-t^2{\tan }^2\alpha }$
$(t_1+t_2{)}_{\text{at}t=1}=\frac{2(3{\tan }^2\alpha +10}{1-{\tan }^2\alpha }$
Range will be $(-\infty ,-6)\cup [2,\infty )$