Question

The area (in sq. units) of the triangle formed by the tangent and normal to the ellipse \( 9x^2 + 4y^2 = 72 \) at the point (2, 3) with the X-axis is

• A. \(\frac{25}{2}\)
• B. \(\frac{39}{4}\)
• C. \(\frac{35}{4}\)
• D. \(\frac{45}{4}\)

Hint:

For an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), tangent at \((x_1, y_1)\): \(\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1\). Normal: \(\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2\).

Answer 1

The Correct Option is B

Rewrite the ellipse \( 9x^2 + 4y^2 = 72 \) in standard form: \[ \frac{x^2}{8} + \frac{y^2}{18} = 1 \implies a^2 = 8, b^2 = 18 \implies a = 2\sqrt{2}, b = 3\sqrt{2} \] The tangent to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) at \((x_1, y_1)\) is: \[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \] At \((2, 3)\), with \(a^2 = 8\), \(b^2 = 18\): \[ \frac{2x}{8} + \frac{3y}{18} = 1 \implies \frac{x}{4} + \frac{y}{6} = 1 \implies 3x + 2y = 12 \] X-axis intercept (\(y = 0\)): \(3x = 12 \implies x = 4\). Point: \((4, 0)\). The normal at \((x_1, y_1)\) is: \[ \frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2 \] At \((2, 3)\), with \(a^2 = 8\), \(b^2 = 18\): \[ \frac{8x}{2} - \frac{18y}{3} = 8 - 18 \implies 4x - 6y = -10 \implies 2x - 3y = -5 \] X-axis intercept (\(y = 0\)): \(2x = -5 \implies x = -\frac{5}{2}\). Point: \(\left(-\frac{5}{2}, 0\right)\). The triangle vertices are \((4, 0)\), \(\left(-\frac{5}{2}, 0\right)\), and \((2, 3)\). Base length (distance along X-axis): \[ \left| 4 - \left(-\frac{5}{2}\right) \right| = 4 + \frac{5}{2} = \frac{13}{2} \] Height is the y-coordinate of \((2, 3)\): 3. Area: \[ \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{13}{2} \times 3 = \frac{39}{4} \] Option (2) is correct. Options (1), (3), and (4) do not match the calculated area.