Find pints on the curve `(x^2)/9+(y^2)/(16)=1` at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis.

Explanation

Equation of curve
`x^(2)/9 +y^(2)/16 = 1`…(1)
`rArr (2x)/9+(2y)/16(dy)/(dx) = 0`
` rArr (dy)/(dx) = - (16x)/(9y)` ….(2)
(i) Tangent is parallel to x-axis
`rArr (dy)/(dx) = 0`
` rArr (-16 x)/(9y) = 0`
` rArr x= 0`
put x = 0 in equation (1)
` 0+(y^(2))/16 = 1`
` rArr y^(2) = 16`
` rArr y = pm 4`
`:. ` The tangents drawn at points (0, 4) and (0, -4) of the curve are parallel to x-axis.
(ii) Tangent is parallel to y-axis
` rArr (dx)/(dy) = 0`
` rArr (-9y)/(16x) = 0`
` rArr y= 0`
put y = 0 in equation (1)
` x^(2)/9 + 0 = 1 `
` rArr x^(2) = 9`
` rArr x = pm 3`
`:. ` The tangents drawn at points (3, 0) and (-3, 0) are parallel to y-axis.