Question

The range of $\lambda$ if the point $(\lambda ,\lambda +1)$ lies inside the parabola $y^2=14x$

• A. $(\frac{12-\sqrt{140}}{2},\frac{12+\sqrt{140}}{2})$
• B. $(\frac{14-\sqrt{190}}{2},\frac{14+\sqrt{190}}{2})$
• C. $(6,8)$
• D. $(\frac{12-\sqrt{140}}{2},\frac{14+\sqrt{190}}{2})$

Answer 1

The correct option is A

$(\frac{12-\sqrt{140}}{2},\frac{12+\sqrt{140}}{2})$

The point $(x_1,y_1)$ lies inside $y^2-14x=0$ if
$y_1^2-14x_1<0$

$\Rightarrow (\lambda +1{)}^2-14\lambda <0$

$\Rightarrow {\lambda }^2+2\lambda +1-14\lambda$

$\Rightarrow {\lambda }^2-12\lambda +1<0$

$\lambda \in (\frac{12-\sqrt{{12}^2-4}}{2},\frac{12+\sqrt{{12}^2-4}}{2})$

$=(\frac{12-\sqrt{140}}{2},\frac{12+\sqrt{140}}{2})$