Pythagoras theorem

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we have L(BC)= L(AB)=X+Y & we want to get L(AC)= (2X+Y)

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If y = f(x) be a given function, then the differential coefficient f' (x) or dy at the point P (x , y ) is (i) If the tangent at P (x1,y1) of the curve y = f(x) is parallel to the x- axis (or perpendicular to y- axis) then  = 0 i.e. its slope will be zero. FGdy J dx 1 1 m = H K = 0 the trigonometrical tangent of the angle  (say) which the positive direction of the tangent to the curve at P makes with the positive direction of x- axis Gdy J, therefore represents the slope of the tangent. Thus dx (x1,y1) The converse is also true. Hence the tangent at (x1,y1) is parallel to x- axis.  GJ = 0 (x1,y1) (ii) If the tangent at P (x , y ) of the curve y = 1 1 f (x) is parallel to y - axis (or perpendicular to x-axis) then  =  / 2 , and its slope will be infinity i.e. dy m = =  dx (x ,y ) The converse is also true. Hence the tangent at (x1, y1) is parallel to y- axis Fdy (x1,y1) Thus (i) The inclination of tangent with x- axis. dy (iii) If at any point P (x1, y1) of the curve y = f(x), the tangent makes equal angles with the axes, then at the point P,  =  / 4 or 3 / 4 , = tan–1 GHdxJK Hence at P, tan  = dy/dx =  1. The (ii) Slope of tangent = dy dx (iii) Slope of the normal = – dx/dy Ex.1 Find the following for the curve y2 = 4x at point (2,–2) (i) Inclination of the tangent (ii) Slope of the tangent (iii) Slope of the normal Sol. Differentiating the given equation of curve, we get dy/dx = 2/y = –1 at (2,–2) so at the given point. (i) Inclination of the tangent = tan–1(–1) = 135º (ii) Slope of the tangent = –1 (iii) Slope of the normal = 1 converse of the result is also true. thus at (x1,y1) the tangent line makes equal angles with the axes.  GJ =  1 (x1,y1) Ex.2 The equation of tangent to the curve y2 = 6x at (2, – 3). (A) x + y – 1 = 0 (B) x + y + 1 = 0 (C) x – y + 1 = 0 (D) x + y + 2 = 0 Sol. Differentiating equation of the curve with respect to x (a) Equation of tangent to the curve y = f(x) at A (x1,y1) is 2y dy = 6  dx FGdy J dx (2,3) = 3 = –1 3 y – y1 = FGdy J (x1,y1) (x–x1) Therefore equation of tangent is y + 3 = – (x – 2)  x + y + 1 = 0 Ans. [B] Ex.3 The equation of tangent at any of the curve x = at2, y = 2at is - (A) x = ty + at2 (B) ty + x + at2 = 0 (C) ty = x + at2 (D) ty = x + at3 2 a 1 Sol. dy/dx = (dy/dt)/(dx/dt) = 2 at = t  equation of the tangent at (x,y) point is (y – 2 at) = 1 (x – at2) t  ty = x + at2 Ans.[C] Ex.4 The equation of the tangent to the curve x2 (x – y) + a2 (x + y) = 0 at origin is- (A) x + y + 1 = 0 (B) x + y + 2 = 0 (C) x + y = 0 (D) 2x – y = 0 Sol. Differentiating equation of the curve w.r.t. x dy/dx = – y x (i) If tangent line is parallel to x - axis, then dy/dx = 0  y = 0 and x = a Thus the point is (a,0) (ii) If tangent is parallel to y – axis , then dy/dx =   x = 0 and y = a Thus the point is (0,a) (iii) If tangent line makes equal angles with both axis , then dy/dx =  1  y =

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