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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  
Given curve is, The slope of the tangent at x = 3 is given by Hence, the slope of tangent at point x = 3 is 24
Given curve is, The slope of the tangent at x = 10 is given by  at x = 10                         hence, slope of tangent at x = 10 is 
Given curve is, The slope of the tangent at x = 2 is given by Hence, the slope of the tangent at point x = 2 is 11
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