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Question: The pole of the line 2x + y + 5 = 0 w.r.t. the ellipse \(\frac{x^{2}}{4} + \frac{y^{2}}{9}\) = 1 is...

The pole of the line 2x + y + 5 = 0 w.r.t. the ellipse

x24+y29\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1 is

A

(85,95)\left( \frac{8}{5},\frac{9}{5} \right)

B

(85,95)\left( - \frac{8}{5},\frac{9}{5} \right)

C

(85,95)\left( \frac{8}{5}, - \frac{9}{5} \right)

D

(85,95)\left( \frac{- 8}{5},\frac{- 9}{5} \right)

Answer

(85,95)\left( \frac{- 8}{5},\frac{- 9}{5} \right)

Explanation

Solution

The pole of the line lx+my+n = 0 w.r.to x2a2+y2b2=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 is

(a2ln,6mub2mn)\left( \frac{- a^{2}\mathcal{l}}{n},\mspace{6mu}\frac{- b^{2}m}{n} \right) =(85,95)\left( \frac{- 8}{5},\frac{- 9}{5} \right)