Question

An ellipse with foci $(2,2)$, $(3,-5)$ passes through $(6,-1)$, then its semi-latus rectum is:

• A. $\frac{7}{2}$
• B. $\frac{5}{2}$
• C. $\frac{9}{2}$
• D. $\frac{11}{2}$

Answer 1

Answer: (B)

$\frac{5}{2}$

Distance $S_{1}P=\sqrt{{(6-3)}^2+{(5-1)}^2}$

$\therefore S_{1}P=5$

Distance $S_{2}P=\sqrt{{(6-2)}^2+{(2+1)}^2}$

$\therefore S_{2}P=5$

and we know $S_{1}P+S_{2}P=2a$

$\therefore a=5$

Distance $S_1{S}_2=\sqrt{{(3-2)}^2+{(2+5)}^2}$

and $S_1{S}_2=2ae$

$\therefore ae=\frac{\sqrt{50}}{2}$

and, $b^2=a^2-{\left(ae\right)}^2$

or, $b^2=25-\frac{50}{4}$

$\therefore b^2=\frac{25}{2}$

Length of semi-latus rectum $=\frac{b^2}{a}$

$\therefore$ Length of semi latus rectum $=\frac{5}{2}$