Question

Tangent to the ellipse $\frac{x^2}{32}+\frac{y^2}{18}=1$ having slope $\frac{-3}{4}$ meets the co-ordinate axes in A and B. Find the area of the triangle AOB, where O is the origin

• A. 12 units
• B. 8 units
• C. 24 units
• D. 32 units.

Answer 1

Answer: (C)

24 units

Answer 2

Tangent to the ellipse$\frac{x^{2}}{a^{2}} +\frac{y^{2}}{b^{2}} = 1$ with slope $m$ is $y= mx+\sqrt{a^{2}m^{2}+b^{2}}$ Here $a^{2}= 32, b^{2}=18, m= - 3/4$ $\therefore$ tangent is $y= -\frac{3}{4}x +\sqrt{32\cdot\frac{9}{16}+18}$ $= -\frac{3}{4}x+6$ $\therefore 3x+4y=24$ $\Rightarrow \frac{x}{8}+\frac{y}{6} = 1$ $\therefore A$ is $\left(8, 0\right) B$ is $\left(0, 6\right)$ $\therefore$ area of $\Delta AOB = \frac{1}{2} \left(8\right)\left(6\right)$ $= 24$ sq units