Coordinate Geometry
Common tangent to circle and ellipse; area of triangle
MJMT_Full_Test_02
Grade 12
Question:
Circle $x^2+y^2=r^2$ meets ellipse $16x^2+25y^2=400$; $4<r<5$. Common tangent (slope $m>0$) in 2nd quadrant meets axes at $P$, $Q$. Area of $\triangle OPQ$ (O=origin) is minimum. Then $m$ is
$\dfrac{2}{\sqrt5}$
$\dfrac{3}{5}$
$\dfrac{4}{5}$
$\dfrac{5}{3}$
Step-by-Step Solution
Key Concept: Common tangent to both: $y=mx\pm\sqrt{25m^2+16}$ (ellipse tangent) must be tangent to circle $\Rightarrow r^2=25m^2+16$. Area of $\triangle OPQ=\frac{1}{2}\cdot|P|\cdot|Q|=\frac{(25m^2+16)}{2m}$. Minimize.
$m=4/5$.
Correct Answer: 3