Find points on the curve at which the tangents are
(i) parallel to x-axis (ii) parallel to y-axis.
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Answer:
The equation of the given curve is.
x^2/9+y^2/16=1
On differentiating both sides with respect to x, we have:
2x/9+2y/16*dy/dx=0
dx/dy=-16x/9y
(i) The tangent is parallel to the x-axis if the slope of the tangent is i.e., 0
-16x/9y=0
which is possible if x = 0.
then,x^2/9+y^2/16=1
Then,for x = 0
y^2=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⇒ y = 0.
-1/(-16x/9y)=9y/16x=0
y=0
then ,x^2/9+y^2/16=1
,for y =0
x=. ± 3
Hence, the points at which the tangents are parallel to the y-axis are
(3, 0) and (− 3, 0).
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