Question

Find the range of values of $a$ for which the point $(a^2-1,a)$ lies inside the parabola $y^2=8x$.

• A. $(-\infty ,-\sqrt{\frac{4}{7}})\cup (\sqrt{\frac{8}{7}},\infty )$
• B. $(-\infty ,-\sqrt{\frac{4}{7}})\cup (\sqrt{\frac{4}{7}},\infty )$
• C. $(-\infty ,-\sqrt{\frac{8}{7}})\cup (\sqrt{\frac{4}{7}},\infty )$
• D. $(-\infty ,-\sqrt{\frac{8}{7}})\cup (\sqrt{\frac{8}{7}},\infty )$

Answer 1

The correct option is D $(-\infty ,-\sqrt{\frac{8}{7}})\cup (\sqrt{\frac{8}{7}},\infty )$

For point $(a^2-1,a)$ to lie inside the parabola $y^2=8x$
$a^2-8(a^2-1)<0$
$\Rightarrow a^2-\frac{8}{7}>0$
$\Rightarrow (a-\sqrt{\frac{8}{7}})(a+\sqrt{\frac{8}{7}})>0$
$\therefore a\in (-\infty ,-\sqrt{\frac{8}{7}})\cup (\sqrt{\frac{8}{7}},\infty )$