Question

The pole of the line y = x+2 with respect to the ellipse x² + 2y² - 4x + 12y + 14 = 0 is

• A. $\left( \frac{9}{4},\mspace{6mu}\frac{- 5}{24} \right)$
• B. $\left( \frac{6}{7},\mspace{6mu}\frac{- 17}{7} \right)$
• C. $\left( - 26,\mspace{6mu}\frac{35}{4} \right)$
• D. $\left( \frac{5}{17},\mspace{6mu}\frac{- 17}{7} \right)$

Answer 1

Answer: (B)

$\left( \frac{6}{7},\mspace{6mu}\frac{- 17}{7} \right)$

Answer 2

Let (x₁ y₁) be the pole, then equation of polar is S₁ = 0.

$\mathbf{x}\mathbf{x}_{\mathbf{1}}\mathbf{+ 2y}\mathbf{y}_{\mathbf{1}}\mathbf{-}\mathbf{2(x +}\mathbf{x}_{\mathbf{1}}\mathbf{) + 6(y +}\mathbf{y}_{\mathbf{1}}\mathbf{) + 14 = 0}$

compare the above line with x – y + 2 =0

then $\frac{x_{1} - 2}{1} = \frac{2y_{1} + 6}{- 1} = \frac{- 2x_{1} + 6y_{1} + 14}{2}$ ⇒ (x₁, y₁)=

(6/7, -17/7)