If the tangent at P(1,1) on y²=x(2-x)² meets the curve again at Q, then Q is

Correct Answer is : (94,38)≤ft(9/4, 3/8)(49,83)

Correct Answer is : (94,38)≤ft(9/4, 3/8)(49,83)

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BITSAT 2021 Paper

Section: Mathematics

Question : 141 of 150

Marks: +1, -0

If the tangent at

P(1,1)

y^{2}=x(2-x)^{2}

Q

Q

Solution:

y^{2}=x(2-x)^{2} \Rightarrow y^{2}=x^{3}-4 x^{2}+4 x

...(i)

⇒

2 y \;\frac{dy}{dx}=3 x^{2}-8 x+4

⇒

\frac{dy}{dx}=\frac{3 x^{2}-8 x+4}{2 y}

⇒

\left[\frac{dy}{dx}\right]_{P}=\frac{3-8+4}{2}=-\frac{1}{2}

∴ Equation of tangent at

P

is:

y-1=-\frac{1}{2}(x-1)

⇒

x+2 y-3=0

Using

y=\frac{3-x}{2}

in (i), we get:

\left(\frac{3-x}{2}\right)^{2}

= x^{3}-4 x^{2}+4 x

⇒

4 x^{3}-17 x^{2}+22 x-9=0

..(ii)

which has two roots 1,1(Because of (ii) being tangent at

(1,1)

Sum of 3 roots

=\frac{17}{4}

\therefore 3\text{rd}

root

=\frac{17}{4}-2

=\frac{9}{4}

Then,

y=\frac{3-\frac{9}{4}}{2}=\frac{3}{8}

∴

Q

\left(\frac{9}{4}, \frac{3}{8}\right)

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