Ellipse and Parabola — Common Tangents
DAILY_CHALLENGE
Grade None
Question:
Let $T_1$ and $T_2$ be two distinct common tangents to the ellipse $E:\dfrac{x^2}{6}+\dfrac{y^2}{3}=1$ and the parabola $P:y^2=12x$. Suppose that the tangent $T_1$ touches $P$ and $E$ at the points $A_1$ and $A_2$, respectively and the tangent $T_2$ touches $P$ and $E$ at the points $A_4$ and $A_3$, respectively. Then which of the following statements is(are) true?
The area of the quadrilateral $A_1A_2A_3A_4$ is 35 square units
The area of the quadrilateral $A_1A_2A_3A_4$ is 36 square units
The tangents $T_1$ and $T_2$ meet the $x$-axis at the point $(-3,0)$
The tangents $T_1$ and $T_2$ meet the $x$-axis at the point $(-6,0)$
Step-by-Step Solution
Key Concept: Common tangent to parabola y=mx+a/m and ellipse: match the condition c²=a²m²+b²
Tangent to $y^2=12x$: $y=mx+3/m$. Tangent to $x^2/6+y^2/3=1$: $y=mx\pm\sqrt{6m^2+3}$.
For common tangent: $3/m=\pm\sqrt{6m^2+3}\Rightarrow9/m^2=6m^2+3\Rightarrow6m^4+3m^2-9=0\Rightarrow(2m^2+3)(m^2-1)=0\Rightarrow m=\pm1$.
$T_1$ ($m=1$): $y=x+3$. Touches $P$ at: $(x+3)^2=12x\Rightarrow(x-3)^2=0\Rightarrow A_1=(3,6)$. Touches $E$ at: $x^2/6+(x+3)^2/3=1\Rightarrow(x+2)^2=0\Rightarrow A_2=(-2,1)$.
$T_2$ ($m=-1$): $y=-x-3$. Similarly $A_4=(3,-6)$, $A_3=(-2,-1)$.
Area of $A_1A_2A_3A_4$: vertices $(3,6),(-2,1),(-2,-1),(3,-6)$. This is a trapezoid with parallel sides $A_2A_3$ (length 2, at $x=-2$) and $A_1A_4$ (length 12, at $x=3$), width 5.
Area $=\dfrac{1}{2}(2+12)\times5=35$. ✓ (A TRUE)
$T_1\cap x$-axis: $0=x+3\Rightarrow x=-3$. $T_2$: same point $(-3,0)$. (C TRUE)
Correct Answer: A, C