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Question Number 39582 by Raj Singh last updated on 08/Jul/18

	Answered by MrW3 last updated on 08/Jul/18

	$I t h i n k y o u m e a n t a n g e n t s o f t h e c u r v e$ $a t p o s i t i o n s w h e r e t h e c u r v e c u t s t h e$ $x - a x i s .$ ${y x}^{2} + x^{2} - 5 x + 6 = 0$ ${y x}^{2} + (x - 2) (x - 3) = 0$ $w i t h y = 0 :$ $\Rightarrow x = 2 a n d 3$ $i . e . t h e c u r v e c u t s t h e x - a x i s a t$ $(2, 0) a n d (3, 0) .$ ${y x}^{2} + x^{2} - 5 x + 6 = 0$ $y^{'} x^{2} + 2 x y + 2 x - 5 = 0$ $\Rightarrow y^{'} = \frac{5 - 2 x (1 + y)}{x^{2}}$ $a t (2, 0) :$ $m = y^{'} = \frac{5 - 2 \times 2 (1 + 0)}{2^{2}} = \frac{1}{4}$ $e q . o f t a n g e n t :$ $y - 0 = \frac{1}{4} (x - 2)$ $\Rightarrow x - 4 y = 2$ $a t (3, 0) :$ $m = y^{'} = \frac{5 - 2 \times 3 (1 + 0)}{3^{2}} = - \frac{1}{9}$ $y - 0 = - \frac{1}{9} (x - 3)$ $\Rightarrow x + 9 x = 3$
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