The circle x^2 + y^2 - 8x = 0 and the hyperbo | vedclass

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(C) The directrix intersects the $x$-axis at $(\pm a/e, 0)$.For the nearest directrix,the line $2x + y = 1$ passes through $(a/e, 0)$.Substituting $(a

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$2$

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(C) The directrix intersects the $x$-axis at $(\pm a/e, 0)$.For the nearest directrix,the line $2x + y = 1$ passes through $(a/e, 0)$.Substituting $(a/e, 0)$ into $2x + y = 1$,we get $2(a/e) + 0 = 1$,which implies $2a = e$.For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the condition for the line $y = mx + c$ to be a tangent is $c^2 = a^2m^2 - b^2$.Here,$y = -2x + 1$,so $m = -2$ and $c = 1$.Thus,$1^2 = a^2(-2)^2 - b^2$,which gives $1 = 4a^2 - b^2$,or $b^2 = 4a^2 - 1$.We also know $b^2 = a^2(e^2 - 1)$.Substituting $e = 2a$,we get $b^2 = a^2((2a)^2 - 1) = a^2(4a^2 - 1) = 4a^4 - a^2$.Equating the two expressions for $b^2$: $4a^4 - a^2 = 4a^2 - 1$.$4a^4 - 5a^2 + 1 = 0$.$(4a^2 - 1)(a^2 - 1) = 0$.Since $2a = e$ and $e > 1$,$a > 1/2$. If $a^2 = 1/4$,then $e = 2(1/2) = 1$,which is not possible for a hyperbola.Thus,$a^2 = 1$,which gives $a = 1$.Then $e = 2a = 2(1) = 2$.

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