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Question: If the tangent at \[P(1,1)\] on \({y^2} = x{(2 - x)^2}\) meets the curve again at \(Q\) , then \(Q\)...

If the tangent at P(1,1)P(1,1) on y2=x(2−x)2{y^2} = x{(2 - x)^2} meets the curve again at QQ , then QQ is :
A) (−4,4)( - 4,4)
B) (1,−2)(1, - 2)
C) (94,38)\left( {\dfrac{9}{4},\dfrac{3}{8}} \right)
D) None of these

Explanation

Solution

The derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument . Derivatives are a fundamental rule of calculus. First we find the derivative of the given function and put the given point then we find the slope . After that we find the tangent equation and find the required answer.

Complete step by step answer:
The given equation y2=x(2−x)2{y^2} = x{(2 - x)^2}
⇒y2=x(4−4x+x2)\Rightarrow {y^2} = x(4 - 4x + {x^2})
⇒y2=4x−4x2+x3\Rightarrow {y^2} = 4x - 4{x^2} + {x^3}
First we differentiate both sides of the above equation with respect to xx and we get
⇒2ydydx=4−8x+3x2\Rightarrow 2y\dfrac{{dy}}{{dx}} = 4 - 8x + 3{x^2}
⇒dydx=4−8x+3x22y\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4 - 8x + 3{x^2}}}{{2y}}
Since the point P(1,1)P(1,1) lies on the above equation then it satisfy the equation
Put this value in above equation and we get
⇒dydx∣(1,1)=4−8+32\Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{(1,1)}} = \dfrac{{4 - 8 + 3}}{2}
⇒dydx∣(1,1)=−12\Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{(1,1)}} = - \dfrac{1}{2}
Now we find the equation of tangent of the curve
y−y1=(−12)(x−x1)y - {y_1} = \left( { - \dfrac{1}{2}} \right)(x - {x_1})
Here the point (x1,y1)=(1,1)({x_1},{y_1}) = (1,1) put in the equation of tangent and we get
⇒y−1=(−12)(x−1)\Rightarrow y - 1 = \left( { - \dfrac{1}{2}} \right)(x - 1)
Take cross multiplication and we get
⇒2y−2=−x+1\Rightarrow 2y - 2 = - x + 1
Taking all the functions in one side change the signs , we have
⇒2y+x−2−1=0\Rightarrow 2y + x - 2 - 1 = 0
⇒2y+x−3=0\Rightarrow 2y + x - 3 = 0 …………………………….(1)
Now we put the option one by one and check
We now check (−4,4)( - 4,4) satisfy the above equation (1) or not
Therefore L.H.S. 2×4+(−4)−32 \times 4 + ( - 4) - 3
=8−4−3= 8 - 4 - 3
=1≠R.H.S.= 1 \ne R.H.S.
∴\therefore Option (1) is incorrect .
Now check (1,−2)(1, - 2) satisfy the above equation (1) or not
L.H.S. 2×(−2)+1−32 \times ( - 2) + 1 - 3
=−4+1−3= - 4 + 1 - 3
=−6= - 6 ≠R.H.S. \ne R.H.S.
∴\therefore Option (2) is incorrect .
Now check (94,38)\left( {\dfrac{9}{4},\dfrac{3}{8}} \right) satisfy the above equation (1) or not
L.H.S. 2×38+94−32 \times \dfrac{3}{8} + \dfrac{9}{4} - 3
=34+94−3= \dfrac{3}{4} + \dfrac{9}{4} - 3
=3+9−3×44= \dfrac{{3 + 9 - 3 \times 4}}{4}
=12−124= \dfrac{{12 - 12}}{4}
=0=R.H.S.= 0 = R.H.S.
∴\therefore Option (C) is correct.

Note:

We also solve the problem by using a substitution method . We take the given curve equation and we write the tangent equation and then we substitute the tangent equation in the given curve equation and solve that to get the point QQ .
Only for multiple choice questions we use the above method.