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 equal intercepts, slope $m = -1$. For equal magnitude intercepts, $m = \pm 1$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the area of the triangle formed by the origin and the intercept points of a specific tangent to a given ellipse. Equal intercepts means the slope of the line is $m = -1$.

Step 2: Key Formula or Approach:
1. Standard form of ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
2. Equation of tangent with slope $m$: $y = mx \pm \sqrt{a^2 m^2 + b^2}$.
3. Area of triangle with intercepts $x_0, y_0$: $A = \frac{1}{2} |x_0 y_0|$.

Step 3: Detailed Explanation:
Ellipse: $9x^2 + 16y^2 = 288 \implies \frac{x^2}{32} + \frac{y^2}{18} = 1$.
Here $a^2 = 32$ and $b^2 = 18$.
Equal intercepts implies $m = -1$ (for a line in the first quadrant, $x + y = k$).
Equation of tangent: $y = -1x \pm \sqrt{32(-1)^2 + 18}$
$y = -x \pm \sqrt{50} \implies x + y = \pm 5\sqrt{2}$.
Intercepts on axes are $x_0 = \pm 5\sqrt{2}$ and $y_0 = \pm 5\sqrt{2}$.
Area of $\triangle OAB = \frac{1}{2} \times |5\sqrt{2}| \times |5\sqrt{2}|$
Area $= \frac{1}{2} \times 25 \times 2 = 25$ sq. units.

Step 4: Final Answer:
The area of $\triangle OAB$ is 25 sq. units.