Exams Joint Entrance Examination Mathematics Hyperbola Question 29 Question 29 of 106 Hard NUMERICAL
Consider the hyperbola x 2 a 2 − y 2 b 2 = 1 \frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1 a 2 x 2 − b 2 y 2 = 1 having one of its focus at P ( − 3 , 0 ) P(-3,0) P ( − 3 , 0 ) . If the latus ractum through its other focus subtends a right angle at P and a 2 b 2 = α 2 − β , α , β ∈ N {a}^{2}{b}^{2}=\alpha \sqrt{2}-\beta ,\alpha ,\beta \in N a 2 b 2 = α 2 − β , α , β ∈ N , then α + β \alpha +\beta α + β is \_\_\_\_\_\_\_\_\_ .
Answer: 1944
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a e = 3 , tan 45 ∘ = a e = 3 b 2 a 6 ⇒ b 2 a = 6 … a ( 1 + b 2 a 2 ) = 3 a 2 + b 2 = 9 ⇒ a 2 − 6 a + 9 = 0 ⇒ a = 3 ( 2 − 1 ) ⇒ a 2 b 2 = 9 ( 3 − 2 2 ) ⋅ 6 ⋅ 3 ( 2 − 1 ) = 162 ( 5 2 − 7 ) ⇒ α = 162 × 5 , β = 162 × 7 ⇒ α + β = 162 × 12 = 1944 \begin{gathered} ae=3,\tan {45}^{\circ }=ae=3\frac{\frac{{b}^{2}}{a}}{6}\Rightarrow \frac{{b}^{2}}{a}=6\ldots \, \\ a\sqrt{(1+\frac{{b}^{2}}{{a}^{2}})}=3 \\ {a}^{2}+{b}^{2}=9\, \\ \Rightarrow {a}^{2}-6a+9=0\Rightarrow a=3(\sqrt{2}-1) \\ \Rightarrow {a}^{2}{b}^{2}=9(3-2\sqrt{2})\cdot 6\cdot 3(\sqrt{2}-1) \\ \, =162(5\sqrt{2}-7) \\ \Rightarrow \alpha =162\times 5,\beta =162\times 7 \\ \Rightarrow \alpha +\beta =162\times 12=1944 \end{gathered} a e = 3 , tan 45 ∘ = a e = 3 6 a b 2 ⇒ a b 2 = 6 … a ( 1 + a 2 b 2 ) = 3 a 2 + b 2 = 9 ⇒ a 2 − 6 a + 9 = 0 ⇒ a = 3 ( 2 − 1 ) ⇒ a 2 b 2 = 9 ( 3 − 2 2 ) ⋅ 6 ⋅ 3 ( 2 − 1 ) = 162 ( 5 2 − 7 ) ⇒ α = 162 × 5 , β = 162 × 7 ⇒ α + β = 162 × 12 = 1944 Answer the question to unlock