Question

PSQ is a focal choed of the ellipse $4x^2+9y^2=36$ such that SP= 4. If S is the another focus, write the value of SQ.

Answer 1

The given equation of the ellipse is $4x^2+9y^2=36$ ...(i)
$\frac{x^2}{9}+\frac{y^2}{4}=1$
This is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ where $a^2=9$ and $b^2=4$ i.e., a=3 and b=2
Clearly a>b, therefore the major axis and minor axis of the ellipse are along x axis and y axis respectively.
Let, e be the eccentricity of the ellipse.
Then,
$\therefore e=\sqrt{1-\frac{b^2}{a^2}}$
$e=\sqrt{1-\frac{4}{9}}$
$e=\frac{\sqrt{5}}{3}$
Therefore, coordinates of focus at S i.e., (ae,0)= $(\sqrt{5},0)$ and coordinates of focus at $S^{\prime }=(-ae,0)=(-\sqrt{5},0)$
It is given that PSQ is a focal chord
As we know that,
SP+S'P=2a
$\Rightarrow 4+S^{\prime }P=6$ $[\because SP=4]$
$\Rightarrow S^{\prime }P=2$
Let coordinates of P be (m,n)
As S'P=2
$\therefore \sqrt{(n-0{)}^2+(m+\sqrt{5}{)}^2}=2$
$n^2+m^2+5+2\sqrt{5}m=4$ ...(ii)
and SP=4
$\therefore \sqrt{(n-0{)}^2+(m-\sqrt{5}{)}^2}=4$
$n^2+m^2+5-2\sqrt{5}m=16$ ...(iii)
Subtracting eq. (iii) from eq. (ii), we get
$4\sqrt{5}=-12$
$m=\frac{-3}{\sqrt{5}}$
and we get $n=\frac{4}{\sqrt{5}}$
$\therefore$Coordinates of P are $(\frac{-3}{\sqrt{5}},\frac{4}{\sqrt{5}})$ and coordinates of S are $(\sqrt{5},0)$
Equation of the line segment PS which is extended to PQ is given by
$x+2y=\sqrt{5}$ ...(iv)
Solving eq. (i) and (iv) we get,
$x=\frac{-3}{\sqrt{5}}$ and $y=\frac{4}{\sqrt{5}}$ which are the coordinates of P
and $x=\frac{66\sqrt{5}}{50}$ and $y=\frac{-8\sqrt{5}}{50}$ which would be the coordinates of Q.
$\therefore S^{\prime }Q=\sqrt{{(\frac{-8\sqrt{5}}{50}-0)}^2+(\frac{66\sqrt{5}}{50}+\sqrt{5})}$
$=\frac{26}{5}$