Question

From any point P lying in first quadrant on the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ PN is drawn perpendicular to the major axis such that N lies on major axis. Now PN is produced to the point Q such that NQ equals to PS, where S is a focus. The point Q lies on which of the following lines

• A. 2y-3x-25=0
• B. 3x+5y+25=0
• C. 2x-5y-25=0
• D. 2x-5y+25=0

Answer 1

The correct option is B 3x+5y+25=0


$a^2=25$
$b^2=16$
$e=\sqrt{\frac{25-16}{25}}=\frac{3}{5}$
Let point Q be (h, k), where K < 0
Given that $|K|=a+eh(as{x}_1=h)$
$-y=a+ex$
$-y=5+\frac{3}{5}x$
$3x+5y+25=0$