Question

From any point $P$ lying in the first quadrant on the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1,PN$ is drawn perpendicular to the major axis and produced at $Q$ so that $NQ$ equals to $P{S}^{\prime },$ where $S^{\prime }$ can be any foci. Then the locus of $Q$ can be

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

Answer 1

Answer: (B), (D)

The correct options are
B $3x+5y+25=0$
D $3x-5y=25$


Here $a^2=25$
and
$b^2=16$
$\Rightarrow e=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$

Let point $Q$ be $(h,k)$, where $k<0$
Given that $k=S^{\prime }P=a\pm e{x}_1,$ where $P(x_1,y_1)$ lies on the ellipse

$\Rightarrow |k|=a\pm eh$ $($as $x_1=h)$
$\Rightarrow -y=a\pm ex$
$\Rightarrow 3x+5y+25=0$ or $3x-5y=25$