Question

Find the set of value (s) of ${}^{\prime }{\lambda }^{\prime }$ for which the line $3x-4y+\lambda =0$ intersect the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ at two distinct points.

Answer 1

Solving given line with ellipse, we get $\frac{{(4y-\lambda )}^2}{9\times 16}+\frac{y^2}{9}=1\frac{2y^2}{9}-\frac{y\lambda }{18}+\frac{{\lambda }^2}{144}-1=0$ Since, line intersect the parabola at two distinct points,

$\therefore$ roots of above equation are real & distinct $\therefore D>0\Rightarrow \frac{{\lambda }^2}{{\left(18\right)}^2}-\frac{8}{9}(\frac{{\lambda }^2}{144}-1)>0\Rightarrow -12\sqrt{2}<\lambda <12\sqrt{2}$