Exams Joint Entrance Examination Mathematics Ellipse Question 50 Question 50 of 183 Medium NUMERICAL
Let an ellipse with centre ( 1 , 0 ) (1,0) ( 1 , 0 ) and latus rectum of length 1 2 \frac{1}{2} 2 1 have its major axis along x \mathrm{x} x -axis. If its minor axis subtends an angle 60 ∘ 60^{\circ} 6 0 ∘ at the foci, then the square of the sum of the lengths of its minor an...
Answer: 9
Solutions · 1 Post your solution Official solution
2 b 2 a = 1 2 , tan 30 ∘ = b a e b 2 = a 4 , 1 3 = b 2 a 2 − b 2 ⇒ a 2 − b 2 = 3 b 2 ⇒ b 2 = a 2 4 ⇒ a = 1 , b 2 = 1 4 ⇒ b = 1 2 ⇒ ( 2 a + 2 b ) 2 = 9 \begin{gathered} \frac{2{b}^{2}}{a}=\frac{1}{2},\, \tan {30}^{\circ }=\frac{b}{ae} \\ {b}^{2}=\frac{a}{4},\frac{1}{3}=\frac{{b}^{2}}{{a}^{2}-{b}^{2}}\Rightarrow {a}^{2}-{b}^{2}=3{b}^{2}\Rightarrow {b}^{2}=\frac{{a}^{2}}{4} \\ \Rightarrow \, a=1,{b}^{2}=\frac{1}{4}\Rightarrow b=\frac{1}{2} \\ \Rightarrow \, (2a+2b{)}^{2}=9 \end{gathered} a 2 b 2 = 2 1 , tan 30 ∘ = a e b b 2 = 4 a , 3 1 = a 2 − b 2 b 2 ⇒ a 2 − b 2 = 3 b 2 ⇒ b 2 = 4 a 2 ⇒ a = 1 , b 2 = 4 1 ⇒ b = 2 1 ⇒ ( 2 a + 2 b ) 2 = 9 Answer the question to unlock