Find the points on the  curve 25y​2+6xy+x2-16=0 at which the tangents are,
(i) parallel to  x- axis.
(ii) parallel to y- axis.

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=050ydydx+6xdydx+6y+2x=0dydx50y+6x=-2x-6ydydx=-2x-6y50y+6xand dydxx1,y1=0So, dydxx1,y1=-2x1-6y150y1+6x1=0-2x1-6y1=0-2x1=6y1y1=-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=025x129-2x12+x12=16x12259-2+1=16x12169=16x12=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

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