So I have an ellipse with foci $(1,-1)$ and $(2,-1)$ and $x+y=5$ as tangent at $(m,n)$ . I need to find out the value of $ \frac {1}{e^2} $ where $e$ is the eccentricity of ellipse. Please don't solve the question. My attempt: I tried to write equation of tangent in parametric form and compare coefficients but the number of unknown is more than the number of relations. I need the process to be followed . Thanks in advance.
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Hint:
Let $A=(1,-1)$ and $B=(2,-1)$ and $B'=(x',y')$ be the reflection of $B$ about $x+y=5$.
By $BB' \perp x+y=5$,
$$\frac{y'-(-1)}{x'-2}=1 \tag{1}$$
Equating perpendicular distances, $$\frac{2+(-1)-5}{\sqrt{1^2+1^2}}=-\frac{x'+y'-5}{\sqrt{1^2+1^2}} \tag{2}$$
After solving you could find $2a=AB'$, together with $2ae=AB$.
See also another answer of mine here.
Ng Chung Tak
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Yes, shifting the axes so that the ellipse is centered on the origin is a good way to start. You can then use the fact that the tangents to the shifted ellipse with slope $m$ have equations $y=mx\pm\sqrt{m^2a^2+b^2}$ and the known focal distance to generate a system of equations in $a^2$ and $b^2$, from which the rest should be fairly straightforward.
amd
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