The smallest possible positive slope of a lin | vedclass

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(B) The equation of the ellipse is $9x^2 + 16y^2 = 144$,which can be written as $\frac{x^2}{16} + \frac{y^2}{9} = 1$.Here,$a^2 = 16$ and $b^2 = 9$.The

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$1$

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(B) The equation of the ellipse is $9x^2 + 16y^2 = 144$,which can be written as $\frac{x^2}{16} + \frac{y^2}{9} = 1$.Here,$a^2 = 16$ and $b^2 = 9$.The equation of a line with slope $m$ and $y$-intercept $c$ is $y = mx + c$. Given $c = 5$,the line is $y = mx + 5$.For this line to have a common point with the ellipse,it must be either a secant or a tangent. The condition for the line $y = mx + c$ to be tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$.Substituting the values,we get $5^2 = 16m^2 + 9$.$25 = 16m^2 + 9$ $\Rightarrow 16m^2 = 16$ $\Rightarrow m^2 = 1$.Thus,$m = \pm 1$.The positive slope is $m = 1$. For any slope $m > 1$,the line $y = mx + 5$ will intersect the ellipse at two points,and for $m < 1$,it will not intersect the ellipse. Therefore,the smallest positive slope for which the line has a common point with the ellipse is $1$.

The smallest possible positive slope of a line whose $y$-intercept is $5$ and which has a common point with the ellipse $9x^2 + 16y^2 = 144$ is

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