The sum of the focal distances of any point on the conic (x2/25)+(y2/16)=1 is

Question

The sum of the focal distances of any point on the conic (x2/25)+(y2/16)=1 is (A) 10 (B) 9 (C) 41 (D) 18

Tardigrade Admin · Accepted Answer

Comparing the given equation, (x2/25)+(y2/16)=1 with general equation of ellipse (x2/-a2)+(y2/b2)=1, we get a2=25 ⇒ a=5 We know that, the sum of focal distances =2 a =2 × 5 =10

Q. The sum of the focal distances of any point on the conic $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ is

10

9

41

18

Comparing the given equation,
$\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$
with general equation of ellipse
$\frac{x ^{2}}{- a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , we get
$a^{2} = 25$
$\Rightarrow a = 5$
We know that, the sum of focal distances $= 2 a$
$= 2 \times 5 = 10$

Q. The sum of the focal distances of any point on the conic 25x2​+16y2​=1 is

10

9

41

18

Comparing the given equation, 25x2​+16y2​=1 with general equation of ellipse −a2x2​+b2y2​=1, we get a2=25 ⇒a=5 We know that, the sum of focal distances =2a =2×5=10

Q. The sum of the focal distances of any point on the conic $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ is

Comparing the given equation,
$\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$
with general equation of ellipse
$\frac{x ^{2}}{- a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , we get
$a^{2} = 25$
$\Rightarrow a = 5$
We know that, the sum of focal distances $= 2 a$
$= 2 \times 5 = 10$