Exams Joint Entrance Examination Mathematics Circle Question 8 Question 8 of 219 Hard NUMERICAL
Let the circle C C C touch the line x − y + 1 = 0 x-y+1=0 x − y + 1 = 0 , have the centre on the positive x x x -axis, and cut off a chord of length 4 13 \frac{4}{\sqrt{13}} 13 4 along the line − 3 x + 2 y = 1 -3 x+2 y=1 − 3 x + 2 y = 1 . Let H be the hyperbola x 2 α 2 − y 2 β 2 = 1 \frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1 α 2 x 2 − β 2 y 2 = 1 , who...
Answer: 19
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x − y + 1 = 0 p = r ∣ α − 0 + 1 2 ∣ = r ⇒ ( α + 1 ) 2 = 2 r 2 \begin{gathered} x-y+1=0 \\ p=r \\ |\frac{\alpha -0+1}{\sqrt{2}}|=r\Rightarrow (\alpha +1{)}^{2}=2{r}^{2}\, \end{gathered} x − y + 1 = 0 p = r ∣ 2 α − 0 + 1 ∣ = r ⇒ ( α + 1 ) 2 = 2 r 2
∴ r = 2 2 α e = 32 α = 4 2 α 2 e 2 = 9 ⇒ α = 2 2 ⇒ α 2 = 8 α 2 ( 1 + β 2 α 2 ) = 9 α 2 + β 2 = 9 ∴ β 2 = 1 ∴ 2 α 2 + 3 β 2 = 2 ( 8 ) + 3 ( 1 ) = 19 \begin{gathered} \therefore \, r=2\sqrt{2} \\ \alpha e=32\alpha =4\sqrt{2} \\ {\alpha }^{2}{e}^{2}=9\Rightarrow \alpha =2\sqrt{2}\Rightarrow {\alpha }^{2}=8 \\ {\alpha }^{2}(1+\frac{{\beta }^{2}}{{\alpha }^{2}})=9 \\ {\alpha }^{2}+{\beta }^{2}=9 \\ \therefore {\beta }^{2}=1 \\ \therefore 2{\alpha }^{2}+3{\beta }^{2}=2(8)+3(1)=19 \end{gathered} ∴ r = 2 2 α e = 32 α = 4 2 α 2 e 2 = 9 ⇒ α = 2 2 ⇒ α 2 = 8 α 2 ( 1 + α 2 β 2 ) = 9 α 2 + β 2 = 9 ∴ β 2 = 1 ∴ 2 α 2 + 3 β 2 = 2 ( 8 ) + 3 ( 1 ) = 19 Answer the question to unlock