Question

Find the set of values of $\lambda$ for which the
line $3x-4y+\lambda =0$ intersects the ellipse
$\frac{x^2}{16}+\frac{y^2}{a}=1$ at 2 distinct point.

• A. $(-12,12)$
• B. $(-12\sqrt{2},12\sqrt{2})$
• C. $(12,12\sqrt{2})$
• D. $(-12,12\sqrt{2})$

Answer 1

The correct option is B

$(-12\sqrt{2},12\sqrt{2})$

We know a line intersects an ellipse at 2 distinct points if $c^2$$<$ $a^2{m}^2+b^2$
$\therefore y=\frac{3}{4}x+\frac{\lambda }{4},\frac{x^2}{4^2}+\frac{y^2}{3^2}=1$
$\therefore {\left(\frac{\lambda }{4}\right)}^2<4^2.{\left(\frac{3}{4}\right)}^2+3^2$
$\Rightarrow {\lambda }^2$$<$ $18{\left(4\right)}^2$
$\therefore \left|\lambda \right|$$<$ $12\sqrt{2}$
$\therefore \lambda \in (-12\sqrt{2},12\sqrt{2})$