Dear Student, Solution: Consider the given curvey2=x(2-x)2 Differentiating wrt xddxy2=ddx(x(2-x)2) 2ydydx=(2-x)2+2(2-x)(-1)x⇒2ydydx=4+x2-4x-4x+2x2⇒2ydydx=4+3x2-8x⇒dydx=4+3x2-8x2yNow put (x,y)=(1,1) andSlope=m=4+3(1)2-8(1)2(1)=-12Nowequation of tangent is:y-y1=dydx(x-x1)It is given that (1,) lies on the tangent so,⇒y-1=-12(x-1)⇒y=3-x2Now, put this value of y in the equation of curve,y2=x(2-x)2 ⇒(3-x2)2=x(2-x)2Taking square root on both sides, we get⇒(3-x2)=x(2-x)Let x=t2then,⇒(3-t22)=t(2-t2)⇒3-t2=4t-2t3⇒2t3-t2-4t+3=0Put t=1 and solve2(1)3-(1)2-4(1)+3=5-5=0Now divide the polynomial by (t-1)we get(t-1)(2t2+t-3)=0Factorise (2t2+t-3)=0t=1,1,-32Put the value of t for xx=(-32)2=94For y,y=3-x2⇒y=3-942=12-98=38So, Q=(x,y)=(94,38)It is given that3ab=94 and a2b=38From this, b=4 and a=3Therefore a2+b2=32+42=9+16=25 Regards

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Solution:

Consider the given curve y^{2} = x {(2 - x)}^{2} Differentiating wrt x \frac{d}{dx} y^{2} = \frac{d}{dx} (x {(2 - x)}^{2}) 2 y \frac{dy}{dx} = {(2 - x)}^{2} + 2 (2 - x) (- 1) x \Rightarrow 2 y \frac{dy}{dx} = 4 + x^{2} - 4 x - 4 x + 2 x^{2} \Rightarrow 2 y \frac{dy}{dx} = 4 + 3 x^{2} - 8 x \Rightarrow \frac{dy}{dx} = \frac{4 + 3 x^{2} - 8 x}{2 y} Now put (x, y) = (1, 1) and Slope = m = \frac{4 + 3 {(1)}^{2} - 8 (1)}{2 (1)} = \frac{- 1}{2} Now equation of tangent is : y - y_{1} = \frac{dy}{dx} (x - x_{1}) It is given that (1,) lies on the tangent so, \Rightarrow y - 1 = \frac{- 1}{2} (x - 1) \Rightarrow y = \frac{3 - x}{2} Now, put this value of y in the equation of curve, y^{2} = x {(2 - x)}^{2} \Rightarrow {(\frac{3 - x}{2})}^{2} = x {(2 - x)}^{2} Taking square root on both sides, we get {\Rightarrow (\frac{3 - x}{2}) = \sqrt{x} (2 - x)}_{} Let x = t^{2} then, \Rightarrow (\frac{3 - t^{2}}{2}) = t (2 - t^{2}) \Rightarrow 3 - t^{2} = 4 t - 2 t^{3} \Rightarrow 2 t^{3} - t^{2} - 4 t + 3 = 0 Put t = 1 and solve 2 {(1)}^{3} - {(1)}^{2} - 4 (1) + 3 = 5 - 5 = 0 Now divide the polynomial by (t - 1) we get (t - 1) (2 t^{2} + t - 3) = 0 Factorise (2 t^{2} + t - 3) = 0 t = 1, 1, \frac{- 3}{2} Put the value of t for x x = {(\frac{- 3}{2})}^{2} = \frac{9}{4} For y, y = \frac{3 - x}{2} \Rightarrow y = \frac{3 - \frac{9}{4}}{2} = \frac{12 - 9}{8} = \frac{3}{8} So, Q = (x, y) = (\frac{9}{4}, \frac{3}{8}) It is given that \frac{3 a}{b} = \frac{9}{4} and \frac{a}{2 b} = \frac{3}{8} From this, b = 4 and a = 3 Therefore a^{2} + b^{2} = 3^{2} + 4^{2} = 9 + 16 = 25

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