Ellipse
5.0 Focal Distance of a point
5.0 Focal Distance of a point
The sum of focal distances of any point on the ellipse is equal to the major axis.
Proof: Let us assume the equation of ellipse be $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$.
As shown in figure $13$, the foci of ellipse are $S(ae,0)$ and $S'(-ae,0)$ and equation of directrices $MZ$ and $M'Z'$ are $x = \frac{a}{e}$ and $x = - \frac{a}{e}$. Let $P(x,y)$ be any point on ellipse.
From the definition of ellipse, $SP = e \times PM$ and ${S'}P = e \times P{M'}$
or,
$$SP = e(\frac{a}{e} - {x_1}) = a - e{x_1}...(1)$$ and
$$S'P = e(\frac{a}{e} + {x_1})= a + e{x_1}...(2)$$
Adding equation $(1)$ and $(2)$, we get
$$SP+S'P=a - e{x_1}+a + e{x_1}$$
$$SP+S'P=2a= length\ of\ major\ axis$$
Another definition of Ellipse: An ellipse is the locus of a point which moves in a plane such that the sum of its distances from two points in the same plane is always constant.
Question 5. Find the eccentricity of an ellipse having foci $S(3,1)$ and $S'(1,1)$ and passes through the point $P(1,3)$.
Solution: As we know that, $$SP+S'P=2a$$
Using distance formulae, we get
$$\begin{equation} \begin{aligned} \sqrt {{{(3 - 1)}^2} + {{(1 - 3)}^2}} + \sqrt {{{(1 - 1)}^2} + {{(1 - 3)}^2}} = 2a \\ 2\sqrt 2 + 2 = 2a \\ a = \sqrt 2 + 1 \\\end{aligned} \end{equation} $$
Distance between foci $S$ and $S'$=$2ae$
$$\begin{equation} \begin{aligned} \sqrt 4 = 2ae \\ \frac{1}{a} = e \\ e = \frac{1}{{\sqrt 2 + 1}} \\\end{aligned} \end{equation} $$
