Find points on the curve at which the tangents are (i) parallel to x -axis (ii) parallel to y -axis
The equation of the given curve is,
x 2 9 + y 2 16 = 1
The slope of the tangent at any point ( x , y ) is given as,
slope = d y d x
The slope of the given curve is given by differentiating with respect to x .
2 x 9 + 2 y 16 d y d x = 0 d y d x = − 16 x 9 y
When the tangent is parallel to the x-axis then slope of the tangent must be 0 .
− 16 x 9 y = 0
The above equation is possible when x = 0 .
The coordinate of y when x = 0 is given as,
0 + y 2 16 = 1 y 2 = 16 y = ± 4
Therefore, the points where tangent is parallel to x-axis are ( 0 , 4 ) and ( 0 , − 4 ) .
When the tangent is parallel to x-axis then slope of the normal must be 0 .
The slope of the normal is given as,
Slope of normal = − 1 Slope of the tangent
Therefore, slope of the normal is,
− 1 ( − 16 x 9 y ) = 9 y 16 x
Slope of the normal must be 0 , so
9 y 16 x = 0
The above equation is valid for y = 0 .
The coordinate of x when y = 0 is given as,
x 2 9 = 1 x 2 = 9 x = ± 3
Thus, the points where tangent is parallel to y-axis are ( 3 , 0 ) and ( − 3 , 0 ) .
The equation of the given curve is,
The slope of the tangent at any point
The slope of the given curve is given by differentiating with respect to
When the tangent is parallel to the x-axis then slope of the tangent must be
The above equation is possible when
The coordinate of
Therefore, the points where tangent is parallel to x-axis are
When the tangent is parallel to x-axis then slope of the normal must be
The slope of the normal is given as,
Therefore, slope of the normal is,
Slope of the normal must be
The above equation is valid for
The coordinate of
Thus, the points where tangent is parallel to y-axis are
at which the tangents are