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: Tangency Condition:
Line $y=mx+c$ touches $\dfrac{x^2}{4}+\dfrac{y^2}{3}=1$ when $c^2=4m^2+3$.

Step 2: Substitute:
$m=-\tfrac12$ gives $c^2=1+3=4$, so $c=\pm2$.

Step 3: Triangle:
The line $y=-\tfrac12x+2$ has y-intercept 2 and x-intercept $2\div\tfrac12=4$. Area $=\tfrac12(4)(2)=4$. Option (A).

Final Answer:
Option (A). \[ \boxed{\text{(A) } 4\ \text{sq. units}} \]