Find the points on the curve 25y2+6xy+x2-16=0 at which the tangents are, (i) parallel to x- axis. (ii) parallel to y- axis. Share with your friends Share 7 Neha Sethi answered this Dear student i) The equation of the given curve is, 25y2+6xy+x2-16=0Let the required point be Px1,y1Since the tangent to the curve at Px1,y1 is parallel to x axis, thendydxx1,y1=0Computing dydx.We have , 25y2+6xy+x2-16=0differentiate wrt x, we get252ydydx+6xdydx+y+2x=0⇒50ydydx+6xdydx+6y+2x=0⇒dydx50y+6x=-2x-6y⇒dydx=-2x-6y50y+6xand dydxx1,y1=0So, dydxx1,y1=-2x1-6y150y1+6x1=0⇒-2x1-6y1=0⇒-2x1=6y1⇒y1=-x13Since Px1,y1 lies on the given curve , so we have25y12+6x1y1+x12-16=0Putting the value of y, we get25-x132+6x1-x13+x12-16=0⇒25x129-2x12+x12=16⇒x12259-2+1=16⇒x12169=16⇒x12=9So, x1=±3So, when x1=3 ,y1=-33=-1and when x1=-3 ,y1=--33=1So required points are3,-1 and -3,1 Similarly try other part. Regards 0 View Full Answer
