From any point P lying in the first quadrant on the ellipse (x2/25)+(y2/16)=1, P N is drawn perpendicular to the major axis and produced at Q so that NQ equals to PS, where S is a focus. Then the locus of Q is

Question

From any point P lying in the first quadrant on the ellipse (x2/25)+(y2/16)=1, P N is drawn perpendicular to the major axis and produced at Q so that NQ equals to PS, where S is a focus. Then the locus of Q is (A) 5 y-3 x-25=0 (B) 3 x+5 y+25=0 (C) 3 x-5 y-25=0 (D) none of these

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/1e1a6bc82da0a5c73775326b80347101-.png" /> a2=25 and b2=16 or e =√1-(16/25)=(3/5) Let point Q be (h, k), where k < 0. Given that k = SP = a +ex1, where P ( x 1, y 1) lies on the ellipse, Then, | k |= a + eh ( text As x 1= h ) or -y=a+e x or 3 x+5 y+25=0

Q. 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 $PS$ , where $S$ is a focus. Then the locus of $Q$ is

$5 y - 3 x - 25 = 0$

$3 x + 5 y + 25 = 0$

$3 x - 5 y - 25 = 0$

none of these

$a^{2} = 25$ and $b^{2} = 16$
or $e = 1 - \frac{16}{25} = \frac{3}{5}$
Let point Q be (h, k), where k $< 0$ .
Given that $k = SP = a + e x_{1}$ ,
where $P (x_{1}, y_{1})$ lies on the ellipse,
Then,
$∣ k ∣ = a + e h (As x_{1} = h)$
or $- y = a + e x$
or $3 x + 5 y + 25 = 0$

Q. From any point P lying in the first quadrant on the ellipse 25x2​+16y2​=1,PN is drawn perpendicular to the major axis and produced at Q so that NQ equals to PS, where S is a focus. Then the locus of Q is

5y−3x−25=0

3x+5y+25=0

3x−5y−25=0