Question

If a tangent having slope of – $\frac{4}{3}$ 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 D 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.

\ $\frac{x_{1}^{2}}{18}$ + $\frac{y_{1}^{2}}{32}$ = 1 (i)

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 $\left( 0,\frac{32}{y_{1}} \right)$.

If is given that the slope of the tangent at (x₁, y₁) is – $\frac{4}{3}$. Therefore

– $\frac{x_{1}}{18}$· $\frac{32}{y_{1}}$ = – $\frac{4}{3}$ Ž $\frac{x_{1}}{y_{1}}$ = $\frac{3}{4}$ Ž $\frac{x_{1}}{3}$ = $\frac{y_{1}}{4}$ = k (say)

\ x₁ = 3k, y₁ = 4k.

Putting x₁, y₁ in (i), we get k² = 1.

Now, area of DOAB = $\frac{1}{2}$ 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 (Q k² = 1).