Question

x2/9+y2/16=2 find the radius of curvature​

Answer 1

Step-by-step explanation:

The equation of the given curve is 9x2+16y2=1  

On differentiating both sides with respect to x, we have:

92x+162ydxdy=0

⇒dxdy=9y−16x

(i)

The tangent is parallel to the x-axis if the slope of the tangent is 0. 

i.e.,9y−16x=0, which is possible if x=0.. 

Then, 9x2+16y2=1 

for x=0,  ⇒=y2=16⇒y=±4

Hence, the points at which the tangents are parallel to the x-axis are (0,4) and (0,−4). 

(ii)

The tangent is parallel to the y-axis if the slope of the normal is 0,

which gives (9y−16)−1=16x9y=0 ⇒y=0

Then,