A family of rays having the equation $(3 + 4λ)x + (4 – λ)y – 7 – λ = 0$ is incident on an ellipse $\ E_1$ and the reflected family of rays has the equation $(3 + 4μ)x + (4 – 3μ)y + 7 + μ = 0$. The point source of the incident family of rays and point of convergence of reflected family of rays lie on axis of the ellipse. The foot of perpendicular drawn from $(1, 1)$ to any tangent of the ellipse $(E_1)$ lies on $x^2 + y^2 = 4$. $\ E_2$ is another concentric ellipse that passes through the foci of the first ellipse and also its foci lie on the first ellipse $(E_1)$. What is the length of major axis of $\ E_2$?
My approach $(3 + λ)x + (4 – λ)y – 7 – λ = 0$ represent the equation of line passing through the intersection of the line $3x+4y-7=0$ & $x-y-1=0$ . We get the intersection point as $({11}/{19},~{25}/{19})$.
SIMILARLY FOR $(3 + 4μ)x + (4 – 3μ)y + 7 + μ = 0$ we get intersection point as $(-1,-1)$
Based on my experience on some of the basic problem both the line intersect on the ellipse. One line passes through one focus and the other line passes through another focus. Both lie on axis of ellipse hence $y=0$ in both the line case.
We know that-The locus of the foot of the perpendicular from a focus to a tangent to the ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2}=1 $ is the circle $x^2 + y^2 = a^2$, hence the line passing through $(1,1)$ and foot of perpendicular to the tangent passes through one of the nearest focus,but not able to approach from here onward.