If the straight line $8 x+3 \sqrt{2} y=36$ touches the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=2$ at $(a, b)$, then $a+\sqrt{2} b=$

• A

  $\frac{36}{5 \sqrt{2}}$
• B

  $\frac{8}{3}$
• C

  $\frac{12+2 \sqrt{2}}{3}$
• Correct answer
  D

  $\frac{16}{3}$

Solution

Given, $8 x+3 \sqrt{2} y=36$ $$ \frac{x^2}{9}+\frac{y^2}{4}=2,(a, b) $$
$$ \begin{gathered} \frac{x x_1}{9}+\frac{y y_1}{4}-2=0 \\ 8 x+3 \sqrt{2} y-36=0 \end{gathered} $$ Eqs. (i) and (ii) represents same line, so $$ \begin{aligned} & \frac{x_1}{(9) \cdot(8)}=\frac{y_1}{(4) \cdot(3 \sqrt{2})}=\frac{1}{18} \\ & \Rightarrow \quad x_1=\frac{72}{18}, y_1=\frac{12 \sqrt{2}}{18} \\ & \Rightarrow \quad x_1=4=a \\ & \Rightarrow \quad y_1=\frac{2 \sqrt{2}}{3}=b \\ & \therefore \quad a+\sqrt{2} b=4+\sqrt{2} \cdot \frac{2 \sqrt{2}}{3} \\ & \Rightarrow \quad \frac{12+4}{3}=\frac{16}{3} \\ & \end{aligned} $$

Hence, the correct answer is option: 'D'.