Question

The tangent to the ellipse $9x^{2}+16y^{2}=288$ making equal intercepts on the coordinate axes intersects the X-axis and the Y-axis in the points A and B respectively. Then $A(\triangle OAB)=$ (where O is origin)}

• A. $\frac{25}{2}$ sq. units
• B. 25 sq. units
• C. $\frac{25\sqrt{5}}{2}$ sq. units
• D. $25\sqrt{5}$ sq. units

Hint:

For an ellipse, a tangent with equal intercepts has slope minus one; apply the tangent condition to find its intercepts.

Answer 1

The Correct Option is B

Step 1: Tangent in Slope Form
For $\dfrac{x^2}{32} + \dfrac{y^2}{18} = 1$, the tangent is $y = mx \pm \sqrt{32m^2 + 18}$.

Step 2: Equal Intercepts
Equal intercepts require $m = -1$, so $y = -x \pm\sqrt{50}$. It cuts both axes at distance $\sqrt{50}$ from the origin.

Step 3: Area
The triangle OAB is right angled with both legs $\sqrt{50}$: area $= \dfrac{1}{2}\times 50 = 25$ sq. units, the second option.

Final Answer:
The area is 25 square units. \[ \boxed{\text{(B) }25\ \text{sq. units}} \]