TG EAMCET 2 May 2025 Shift 1 Paper

Let the given focus be $F_1(2,-3)$ and the corresponding directrix be $L_1: 2 x+y-5=0$ and its eccentricity,$e=\;\frac{\sqrt{5}}{3}$Let the other focus be $F_2 (x_2, y_2)$Now, the distance from $F_1(2,-3)$ to $2 x+y-5=0$ is$\;d_1=\;\frac{\left|2(2)+(-3)-5\right|}{\sqrt{2^2+1^2}}$$\;=\;\frac{\left|4-3-5\right|}{\sqrt{4+1}} = \;\frac{\left|-4\right|}{\sqrt{5}} = \;\frac{4}{5}$Let $P(x, y)$ be any point on the ellipse.So, $P F_1 = \sqrt{(x-2)^2+(y+3)^2}$And the distance from $P$ to the directrix $L_1$ is$\;P M_1 = \;\frac{\left|2 x+y-5\right|}{\sqrt{2^2+1^2}}$$\Rightarrow \;\frac{\left|2 x+y-5\right|}{\sqrt{5}}$Since, $P F_1 = e \cdot P M_1$$\; \Rightarrow \sqrt{(x-2)^2+(y+3)^2}$$\; = \;\frac{\sqrt{5}}{3} \cdot \;\frac{\left|2 x+y-5\right|}{\sqrt{5}}$$\; \Rightarrow \;\frac{\left|2 x+y-5\right|}{3}$Squaring on both sides, we get$\;(x-2)^2+(y+3)^2 = \;\frac{(2 x+y-5)^2}{9}$$\; \Rightarrow 9 ((x-2)^2+(y+3)^2)$$\; = 4 x^2+y^2+25+4 x y-20 x-10 y$$\; = 9 x^2-36 x+36+9 y^2+54 y+81$$\; = 4 x^2+y^2+25+4 x y-20 x-10 y$$\; \Rightarrow 5 x^2+8 y^2-4 x y-16 x+64 y+92=0$This is equation of ellipse.Now, slope of directrix is $m_D=-2$ and the line joining the two foci is perpendicular to the directrix.So, $m_F = \;\frac{-1}{m_D} = \;\frac{-1}{-2} = \;\frac{1}{2}$Let, the other focus be $F_2 (x_2, y_2)$So, $\;\frac{y_2-(-3)}{x_2-2} = \;\frac{y_2+3}{x_2-2} = \;\frac{1}{2}$ $\; \Rightarrow 2 (y_2+3) = x_2-2$$\; \Rightarrow \;\; x_2 = 2 y_2 + 8$We know that $c = a e$ and the distance between the focus and directrix is$\;\; \frac{a}{e} - c = \;\frac{a}{e} - a e = \;\frac{a (1-e^2)}{e}$$\; \Rightarrow \;\; \;\frac{a \left(1-\left(\frac{\sqrt{5}}{3}\right)^2\right)}{\frac{\sqrt{5}}{3}} = \;\frac{4}{\sqrt{5}}$$\; \Rightarrow \;\; \;\frac{a \left(1-\frac{5}{9}\right)}{\frac{\sqrt{5}}{3}} = \;\frac{4}{\sqrt{5}} \Rightarrow a \left(\;\frac{4}{9}\right) \times \;\frac{3}{\sqrt{5}} = \;\frac{4}{\sqrt{5}}$$\; \Rightarrow \;\; \;\frac{4 a}{3 \sqrt{5}} = \;\frac{4}{\sqrt{5}} \Rightarrow a=3$$\; \therefore \;\; c = a e = 3 \times \;\frac{\sqrt{5}}{3} = \sqrt{5}$The distance between two foci $= 2 c = 2 \sqrt{5}$Let $F_2 (x_2, y_2)$. So, the distance$\;F_1 F_2 = \sqrt{ (x_2-2)^2 + (y_2+3)^2 } = 2 \sqrt{5}$$\; \Rightarrow \sqrt{ (2 y_2+8-2)^2 + (y_2+3)^2 } = 2 \sqrt{5}$$\; \;\; ( \because x_2=2 y_2+8)$$\; \Rightarrow \;\; \sqrt{4 (y_2+3)^2 + (y_2+3)^2} = 2 \sqrt{5}$$\; \Rightarrow \;\; \sqrt{5 (y_2+3)^2} = 2 \sqrt{5}$$\; \Rightarrow \;\; \sqrt{5} \left| y_2+3 \right| = 2 \sqrt{5} \Rightarrow \left| y_2+3 \right| = 2$So, $y_2+3=2$ or $y_2+3=-2$$\Rightarrow \;\; y_2 = -1 \;\text{ or }\; y_2 = -5$$\therefore \; x_2 = 2(-1)+8 = 6 \;\text{ and }\;$$\; x_2 = 2(-5)+8 = -2$So, $F_2(6,-1)$ and $F_2(-2,-5)$Line $2 x+y-5=0$For $F_1(2,-3), 2(2)+(-3)-5=4-8=-4$For $F_2(6,-1), 2(6)+(-1)-5=12-6=6$Since, -4 and 6 have opposite signs, so foci of an ellipse is $F_2(-2,-5)$.