Question

If the vertices of the ellipse $9x^2+25y^2-18x+100y-116=0$ are the extremities of latus rectum of a parabola whose vertex is $(x_1,y_1)$ where $y_1>-2,$ then

• A. length of latus rectum of the ellipse is $\frac{50}{3}.$
• B. directrix of the parabola is $y-3=0.$
• C. the smaller area enclosed by the parabola and the ellipse is $\frac{5}{6}(9\pi -20)$ sq. units.
• D. the smaller area enclosed by the parabola and the ellipse is $\frac{5}{6}(9\pi -10)$ sq. units.

Answer 1

Answer: (B), (C)

The correct options are
B directrix of the parabola is $y-3=0.$
C the smaller area enclosed by the parabola and the ellipse is $\frac{5}{6}(9\pi -20)$ sq. units.

$9x^2+25y^2-18x+100y-116=0$
$\Rightarrow 9(x-1{)}^2+25(y+2{)}^2=225$
$\Rightarrow \frac{(x-1{)}^2}{25}+\frac{(y+2{)}^2}{9}=1$


Latus rectum $=4a=10$ and focus is $f(1,-2)$
$\because$ distance between focus and directrix $=2a$
So, directrix of parabola $\equiv y=3$

Now, area of $ABCSA$
$=\frac{\text{Area of ellipse}}{2}$
$=\frac{\pi ab}{2}$
$=\frac{\pi \times 5\times 3}{2}$
$=\frac{15}{2}\pi$


Area of $AOCSA=\frac{2}{3}\times \frac{5}{2}\times 10=\frac{50}{3}$
So, required Area
$=\frac{15}{2}\pi -\frac{50}{3}=\frac{5}{6}\times (9\pi -20)$