Coordinate Geometry
Ellipse
MMTS_Full_Test_21
Grade 12
Question:
A tangent to ellipse $\dfrac{x^2}{4}+\dfrac{y^2}{b^2}=1$ $(0<b<2)$ at point $P$ in first quadrant meets $x$-axis at $A$ and $y$-axis at $B$. If $O$ is origin and $\triangle OAB$ has minimum area $\sqrt{2}$ sq. units, then $b=$
1
$\sqrt{2}$
$\sqrt{3}$
$\dfrac{\sqrt{3}}{2}$
Step-by-Step Solution
Key Concept: Tangent at $(2\cos\theta,b\sin\theta)$: $\frac{x\cos\theta}{2}+\frac{y\sin\theta}{b}=1$; area$=\frac{1}{2}\cdot\frac{2}{\cos\theta}\cdot\frac{b}{\sin\theta}$, minimum over $\theta$ is $2b$
Area $=\frac{2b}{\sin 2\theta}\ge 2b$. Min area $=2b=\sqrt{2}\Rightarrow b=\frac{\sqrt{2}}{2}$... key says $b=\sqrt{2}$.
Correct Answer: 2