Ellipse

The normal is 4x sec ϕ - 2 y cosec ϕ = 12 Now, the points Q and M are given by Q = (3 cos ϕ , 0) M = (α , β) Therefore, α = $\frac{3\cos \phi + 4\cos \phi}{2}$ = $\frac{7}{2} \cos \phi$ ⇒ cos ϕ = $\frac{2}{7} \alpha$ and β = sin ϕ ; $\cos^2 \phi + \sin^2 \phi$ = 1. Therefore, $\frac{4}{49} \alpha^2 + \beta^2$ = 1 ⇒ $\frac{4}{49} x^2 + y^2$ = 1 Hence, the rectum is x = ± $2\sqrt{3}$ Hence, $\frac{48}{49} + y^2$ = 1 ⇒ y = ± $\frac{1}{7}$ $\left(\pm 2\sqrt{3}, \pm \frac{1}{7}\right)$ Hence, the locus of M intersects the latus rectum of the given ellipse at the points