MCQ
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
- A$\frac{3}{4}$
- ✓$1$
- C$\frac{4}{3}$
- D$\frac{9}{16}$
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
$y =m x \pm \sqrt{a^2 m^2+b^2}$
$\therefore \quad y =m x \pm \sqrt{16 m^2+9}$
Now, given $y$-intercept $=5$
$\therefore \sqrt{16 m^2+9}=5 \Rightarrow 16 m^2+9=25$
$\Rightarrow \quad 16 m^2=16 \Rightarrow m=\pm 1$
$\therefore \text { Positive slope }=1$
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