Question

A tangent having slope \(-\frac{1}{2}\) to the ellipse \(3x^2+4y^2 = 12\) intersects the X-axis and Y-axis at the points A and B respectively. if O is the origin, then the area of \(△AOB\) is ...

• A. \(4\) sq. units
• B. \(8\) sq. units
• C. \(12\) sq. units
• D. \(16\) sq. units

Hint:

Write the ellipse in standard form and use the tangent y = mx plus or minus root(a^2 m^2 + b^2).

Answer 1

The Correct Option is A

Step 1: Standard Form:
Divide \(3x^2+4y^2=12\) by 12: \(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\), so \(a^2=4\) and \(b^2=3\).

Step 2: Tangent With Slope m:
A tangent with slope \(m\) is \(y=mx\pm\sqrt{a^2m^2+b^2}\). For \(m=-\tfrac12\):
\[ \sqrt{4\cdot\tfrac14+3}=\sqrt4=2 \Rightarrow y=-\frac{x}{2}\pm2 \]

Step 3: Intercepts:
Take \(y=-\tfrac x2+2\). It meets the y-axis at \(B(0,2)\) and the x-axis (set \(y=0\)) at \(A(4,0)\).

Step 4: Area:
\[ \text{Area}=\frac12\times OA\times OB=\frac12\times4\times2=4\ \text{sq. units} \]
The other tangent \(y=-\tfrac x2-2\) gives the same area by symmetry. Options 8, 12 and 16 would need intercept products of 16, 24 and 32.

Final Answer:
The area of the triangle is 4 square units, option (A). \[ \boxed{\text{(A) } 4\ \text{sq. units}} \]