An ellipse in the first quadrant is tangent to co-ordinate axes. If one focus is F 1(3,7), and the other focus is F2(d, 7), then the value of √3 d is equal to

Question

An ellipse in the first quadrant is tangent to co-ordinate axes. If one focus is F 1(3,7), and the other focus is F2(d, 7), then the value of √3 d is equal to (A) 5 (B) 7 (C) 14 (D) 49

Tardigrade Admin · Accepted Answer

We have q 1 q 2=3 d = b 2 and p1 p2=49=b2 Hence 3 d =49 ⇒ √3 d =√49=7

Q. An ellipse in the first quadrant is tangent to co-ordinate axes. If one focus is $F_{1} (3, 7)$ , and the other focus is $F_{2} (d, 7)$ , then the value of $3 d$ is equal to

5

7

14

49

We have $q_{1} q_{2} = 3 d = b^{2}$
and $p_{1} p_{2} = 49 = b^{2}$
Hence $3 d = 49 \Rightarrow 3 d = 49 = 7$

Q. An ellipse in the first quadrant is tangent to co-ordinate axes. If one focus is F1​(3,7), and the other focus is F2​(d,7), then the value of 3d​ is equal to

5

7

14

49

We have q1​q2​=3d=b2 and p1​p2​=49=b2 Hence 3d=49⇒3d​=49​=7

Q. An ellipse in the first quadrant is tangent to co-ordinate axes. If one focus is $F_{1} (3, 7)$ , and the other focus is $F_{2} (d, 7)$ , then the value of $3 d$ is equal to

We have $q_{1} q_{2} = 3 d = b^{2}$
and $p_{1} p_{2} = 49 = b^{2}$
Hence $3 d = 49 \Rightarrow 3 d = 49 = 7$