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BITSAT 2021
Exam
BITSAT
Year
2021
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2021 · Mathematics · Application of Derivatives

If the tangent at P(1,1) on y^ 2 =x(2-x)^ 2 meets the curve again...

If the tangent at P(1,1)P(1,1) on y2=x(2−x)2y^{2}=x(2-x)^{2} meets the curve again at QQ, then QQ is

Options

  1. A(2,2)(2,2)
  2. B(−1,−2)(-1,-2)
  3. C(94,38)\left(\frac{9}{4}, \frac{3}{8}\right)
  4. DNone of these

Correct answer

C. (94,38)\left(\frac{9}{4}, \frac{3}{8}\right)

Step-by-step solution

y2=x(2−x)2⇒y2=x3−4x2+4x… (i) ⇒2ydydx=3x2−8x+4⇒dydx=3x2−8x+42y⇒[dydx]P=3−8+42=−12\begin{array}{l} y^{2}=x(2-x)^{2} \Rightarrow y^{2}=x^{3}-4 x^{2}+4 x \ldots \text{ (i) } \\ \Rightarrow 2 y \frac{\mathrm{dy}}{\mathrm{dx}}=3 x^{2}-8 x+4 \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3 x^{2}-8 x+4}{2 y} \\ \Rightarrow\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{P}}=\frac{3-8+4}{2}=-\frac{1}{2}\end{array}y^{2}=x(2-x)^{2} \Rightarrow y^{2}=x^{3}-4 x^{2}+4 x \ldots(i)⇒2ydydx=3x2−8x+4⇒dydx=3x2−8x+42y⇒[dydx]P=3−8+42=−12(i) \Rightarrow 2 y \frac{\mathrm{dy}}{\mathrm{dx}}=3 x^{2}-8 x+4\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3 x^{2}-8 x+4}{2 y}\Rightarrow\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{P}}=\frac{3-8+4}{2}=-\frac{1}{2}\therefore EquationoftangentatPof tangent at\mathrm{P} is: y−1=−12(x−1)⇒x+2y−3=0⇒x+2y−3=0y-1=-\frac{1}{2}(x-1)\Rightarrow x+2 y-3=0\Rightarrow x+2 y-3=0 Using y=3−x2y=\frac{3-x}{2} in (i),weget:(3−x2)2(i), we get: \left(\frac{3-x}{2}\right)^{2}=x3−4x2+4x⇒4x3−17x2+22x−9=0...(ii) \begin{array}{l} =x^{3}-4 x^{2}+4 x \\ \Rightarrow 4 x^{3}-17 x^{2}+22 x-9=0 \text{...(ii) } \end{array}whichhastworoots1,1(Becauseof(ii)beingtangentatwhich has two roots 1,1 (Because of (ii) being tangent at(1,1)).Sumof3roots=174). Sum of 3 roots =\frac{17}{4}∴3rd root =174−2=94 Then, y=3−942=38∴Q is (94,38)\begin{array}{l} \therefore 3 \mathrm{rd} \text{ root }=\frac{17}{4}-2=\frac{9}{4} \\ \text{ Then, } y=\frac{3-\frac{9}{4}}{2}=\frac{3}{8} \\ \therefore \mathrm{Q} \text{ is }\left(\frac{9}{4}, \frac{3}{8}\right) \end{array}$$
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