We know that Slope of tangent at a point on given curve is given by
Given equation of curve
at point (1,3)
Hence slope of tangent is 2
Now we know that,
Now, equation of tangent at point (1,3) with slope = 2 is
y = 2x + 1
y -2x = 1
Similarly, equation of normal at point (1,3) with slope = -1/2 is
y = mx + c
equation of normal is
We know that Slope of the tangent at a point on the given curve is given by
Given the equation of the curve
at point (0,5)
Hence slope of tangent is -10
Now we know that,
Now, equation of tangent at point (0,5) with slope = -10 is
equation of tangent is
Similarly, the equation of normal at point (0,5) with slope = 1/10 is
equation of normal is
Parallel to y-axis means the slope of the tangent is , means the slope of normal is 0
We know that slope of the tangent at a given point on the given curve is given by
Given the equation of the curve is
Slope of normal =
From this we can say that y = 0
Now. when y = 0,
Hence, the coordinates are (3,0) and (-3,0)
Parallel to x-axis means slope of tangent is 0
We know that slope of tangent at a given point on the given curve is given by
Given the equation of the curve is
From this, we can say that
Now. when ,
Hence, the coordinates are (0,4) and (0,-4)
We know that the slope of the tangent at a point on the given curve is given by
Given the equation of the curve as
It is given thta slope is 0
So,
Now, when x = 1 ,
Hence, the coordinates are
Equation of line passing through and having slope = 0 is
y = mx + c
1/2 = 0 X 1 + c
c = 1/2
Now equation of line is
We know that the slope of the tangent of at the point of the given curve is given by
Given the equation of curve is
It is given that slope is 2
So,
So, this is not possible as our coordinates are imaginary numbers
Hence, no tangent is possible with slope 2 to the curve
We know that the slope of the tangent of at the point of the given curve is given by
Given the equation of curve is
It is given thta slope is -1
So,
Now, when x = 0 ,
and
when x = 2 ,
Hence, the coordinates are (0,-1) and (2,1)
Equation of line passing through (0,-1) and having slope = -1 is
y = mx + c
-1 = 0 X -1 + c
c = -1
Now equation of line is
y = -x -1
y + x + 1 =...
We know that the equation of a line is y = mx + c
Know the given equation of tangent is
y = x - 11
So, by comparing with the standard equation we can say that the slope of the tangent (m) = 1 and value of c if -11
As we know that slope of the tangent at a point on the given curve is given by
Given the equation of curve is
When x = 2 ,
and
When x = -2 ,
Hence, the coordinates...
Points joining the chord is (2,0) and (4,4)
Now, we know that the slope of the curve with given two points is
As it is given that the tangent is parallel to the chord, so their slopes are equal
i.e. slope of the tangent = slope of the chord
Given the equation of the curve is
Now, when
Hence, the coordinates are
We are given :
Differentiating the equation with respect to x, we get :
or
or
It is given that tangent is parallel to the x-axis, so the slope of the tangent is...
The slope of the tangent at a point on given curves is given by
Now,
Similarly,
Hence, the slope of the tangent at is
Now,
Slope of normal = =
Hence, the slope of normal at is
The slope of the tangent at a point on a given curve is given by
Now,
Similarly,
Hence, the slope of the tangent at is -1
Now,
Slope of normal = =
Hence, the slope of normal at is 1