An ellipse has the points (1, –1) and (2, –1) as its foci and x + y – 5 = 0 as one of its tangents. Then the point where this line touches

$$\begin{gathered} Equation of \tan gent is x+y-5=0. \\ Foci of ellipse are (1,-1) and (2,-1). \\ Product of perpendicular from foci upon any \tan gent is equal to square of the semi minor axis. \\ b^2 =\left(\frac{\left|1-1-5\right|}{\sqrt{2}}\right)\left(\frac{\left|2-1-5\right|}{\sqrt{2}}\right) \\ \Rightarrow b^2=\frac{5\times 4}{2}=10 \\ Dis\tan ce between foci =2ae \\ \Rightarrow 2ae=\sqrt{{(2-1)}^2+{(-1+1)}^2} \\ \Rightarrow ae=\frac{1}{2} \\ \Rightarrow a^2{e}^2=\frac{1}{4} \\ \Rightarrow a^2(1-\frac{b^2}{a^2})=\frac{1}{4} \\ \Rightarrow a^2-b^2=\frac{1}{4} \\ \Rightarrow a^2=\frac{41}{4} \\ centre of ellipse=(\frac{3}{2},-1) \\ Equation of ellipse is \\ \frac{{(2x-3)}^2}{41}+\frac{{(y+1)}^2}{10}=1 \\ Substituting x from the equation of \tan gent, we have \\ \frac{{\left(2\right(5-y)-3)}^2}{41}+\frac{{(y+1)}^2}{10}=1 \\ \Rightarrow 10(49+4y^2-28y)+41(y^2+1+2y)=410 \\ \Rightarrow 490+40y^2-280y+41y^2+41+82y=410 \\ \Rightarrow 81y^2-198y+121=0 \\ \Rightarrow {(9y-11)}^2=0 \Rightarrow y=\frac{11}{9} \\ Now,x+y-5=0 \\ \Rightarrow x=5-y \\ \Rightarrow x=5-\frac{11}{9} \\ \Rightarrow x=\frac{34}{9} \end{gathered}$$