The smallest possible positive slope of a lin | vedclass

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Question

(b)

We have equation of ellipse

$9 x^2+16 y^2 =144$

$\Rightarrow \quad \frac{x^2}{16}+\frac{y^2}{9} =1$

Equation of tangent of ellipse is

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(b)

We have equation of ellipse

$9 x^2+16 y^2 =144$

$\Rightarrow \quad \frac{x^2}{16}+\frac{y^2}{9} =1$

Equation of tangent of ellipse is

$y =m x \pm \sqrt{a^2 m^2+b^2}$

$	herefore \quad y =m x \pm \sqrt{16 m^2+9}$

Now, given $y$-intercept $=5$

$	herefore \sqrt{16 m^2+9}=5 \Rightarrow 16 m^2+9=25$

$\Rightarrow \quad 16 m^2=16 \Rightarrow m=\pm 1$

$	herefore 	ext { Positive slope }=1$

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

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