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