Question

$Lety=4x^2$ & $\frac{x^2}{a^2}-\frac{y^2}{16}=1$ intersect

• A.
• B. \left | a \right |> \frac{1}{\sqrt{2}}
• C. \left | a \right |>-\frac{1}{\sqrt{2}}
• D. a>\sqrt{2}

Answer 1

The correct option is A

$y=4x^2\&\frac{1}{4}y=x^2$
Using~$\frac{1}{4a^2}y-\frac{y^2}{16}=1$
$\Rightarrow 4y-a^2{y}^2=16a^2$
$\Rightarrow a^2{y}^2-4y+16a^2=0$
$\Rightarrow D\ge 0$ for intersection of two curves
$\Rightarrow 16-4a^2\left(16a^2\right)\ge 0$
$\Rightarrow 1-4a^4\ge 0$
$\Rightarrow \left(2a^2\right)\le 1$
$\Rightarrow \left|\sqrt{2}a\right|\le 1\Rightarrow -\frac{1}{\sqrt{2}}\le a\le \frac{1}{\sqrt{2}}$