If 2x - y + 1 = 0 is a tangent to the hyperbo | vedclass

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(D) The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.Given the

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$B, C, D$

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(D) The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.Given the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{16} = 1$,we have $b^2 = 16$.The given tangent is $2x - y + 1 = 0$,which can be written as $y = 2x + 1$.Comparing this with $y = mx + c$,we get $m = 2$ and $c = 1$.The condition for tangency is $c^2 = a^2m^2 - b^2$.Substituting the values: $1^2 = a^2(2)^2 - 16$.$1 = 4a^2 - 16 \Rightarrow 4a^2 = 17 \Rightarrow a^2 = \frac{17}{4} \Rightarrow a = \frac{\sqrt{17}}{2}$.Now,we check the sides for a right-angled triangle (where the sum of squares of two smaller sides equals the square of the largest side):$[A]$ $2a = \sqrt{17} \approx 4.12, 4, 1$. Sides: $\sqrt{17}, 4, 1$. $1^2 + 4^2 = 17 = (\sqrt{17})^2$. This forms a right triangle.$[B]$ $2a = \sqrt{17}, 8, 1$. Sides: $\sqrt{17}, 8, 1$. $1^2 + (\sqrt{17})^2 = 18 
eq 8^2 = 64$. Not a right triangle.$[C]$ $a = \frac{\sqrt{17}}{2} \approx 2.06, 4, 1$. Sides: $2.06, 4, 1$. $1^2 + 2.06^2 \approx 5.24 
eq 4^2 = 16$. Not a right triangle.$[D]$ $a = \frac{\sqrt{17}}{2}, 4, 2$. Sides: $2.06, 4, 2$. $2^2 + 2.06^2 \approx 8.24 
eq 4^2 = 16$. Not a right triangle.Thus,options $B, C,$ and $D$ cannot be sides of a right-angled triangle.

If $2x - y + 1 = 0$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{16} = 1$,then which of the following $CANNOT$ be sides of a right-angled triangle?

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