Question

If a tangent having slope of -4/3 to the ellipse $\frac{x^{2}}{18} + \frac{y^{2}}{32} = 1$ intersects the major and minor axes in points A and B respectively, then the area of the ∆OAB is equal to

• A. 12 sq. units
• B. 48 sq. units
• C. 64 sq. units
• D. 24 sq. units

Answer 1

Answer: (D)

24 sq. units

Answer 2

Let P (x₁, y₁) be a point on the ellipse $\frac{x^{2}}{18} + \frac{y^{2}}{32} = 1$.

Then $\frac{x_{1}^{2}}{18} + \frac{y_{1}^{2}}{32} = 1$

The equation of the tangent at (x₁, y₁) is $\frac{xx_{1}}{18} + \frac{yy_{1}}{32} = 1$

This meets the axes at A $\left( \frac{18}{x_{1}},0 \right)$ and B $\left( 0,\frac{32}{y_{1}} \right)$

It is given that the slope of the tangent at (x₁, y₁) is -3/4, therefore,

$\frac{- x_{1}}{18}.\frac{32}{y_{1}} = - \frac{4}{3} \Rightarrow \frac{x_{1}}{y_{1}} = \frac{3}{4}$⇒ $\frac{x_{1}}{3} = \frac{y_{1}}{4} = k$ (say)

⇒ x₁ = 3k, y₁ = 4k

Putting x₁, y₁ in (1), we get, k² = 1

Now, area of ∆OAB = $\frac{1}{2}$. OA.OB = $\frac{1}{2}$.$\frac{18}{x_{1}}.\frac{32}{y_{1}}$

= $\frac{1}{2}\frac{(18)(32)}{x_{1}y_{1}} = \frac{1}{2}\frac{(18)(32)}{(3k)(4k)} = \frac{24}{k^{2}} = 24$