x+3​y=23​ mt​=3​−1​ and point of tangency (233​​,21​) Option (1) x2+9y2=9⇒2x+18yy′=0 ⇒mt​(233​​,21​) i.e.y′=9y−x​=9(21​)−323​​​=3​−1​Option (2) y′=63​x​⇒y′=123​y1​ i.e. mt​=63​1​ Option (3) 2x2−18y2=9⇒4x−36yy′=0 i.e., mt​=9yx​=2.921​33​​=3​1​ Option (4) x2y2=7⇒y′=−yx​ i.e., mt​=−33​ Hence only option (A) is correct