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If the above line touches the ellipse in a single point, then we have:
5 x 2 + 9 ( 2 x + a ) 2 = 1 ;
or 9 x 2 + 5 ( 2 x + a ) 2 = 4 5 ;
or 2 9 x 2 + 2 0 a x + ( 5 a 2 − 4 5 ) = 0 ;
or x = 5 8 − 2 0 a ± 4 0 0 a 2 − 4 ( 2 9 ) ( 5 a 2 − 4 5 ) ;
or x = 5 8 − 2 0 a ± 5 2 2 0 − 1 8 0 a 2 .
In order to have a single point solution for x , we require the discriminant to equal zero: 5 2 2 0 − 1 8 0 a 2 = 0 ⇒ a 2 = 2 9 . Thus, a 2 − 1 0 0 = − 7 1 .