Question

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

Answer 1

Hint: 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 ${y^2} = x{(2 - x)^2}$
$ \Rightarrow {y^2} = x(4 - 4x + {x^2})$
$ \Rightarrow {y^2} = 4x - 4{x^2} + {x^3}$
First we differentiate both sides of the above equation with respect to $x$ and we get
$ \Rightarrow 2y\dfrac{{dy}}{{dx}} = 4 - 8x + 3{x^2}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4 - 8x + 3{x^2}}}{{2y}}$
Since the point $P(1,1)$ lies on the above equation then it satisfy the equation
Put this value in above equation and we get
$ \Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{(1,1)}} = \dfrac{{4 - 8 + 3}}{2}$
$ \Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{(1,1)}} = - \dfrac{1}{2}$
Now we find the equation of tangent of the curve
$y - {y_1} = \left( { - \dfrac{1}{2}} \right)(x - {x_1})$
Here the point $({x_1},{y_1}) = (1,1)$ put in the equation of tangent and we get
$ \Rightarrow y - 1 = \left( { - \dfrac{1}{2}} \right)(x - 1)$
Take cross multiplication and we get
$ \Rightarrow 2y - 2 = - x + 1$
Taking all the functions in one side change the signs , we have
$ \Rightarrow 2y + x - 2 - 1 = 0$
$ \Rightarrow 2y + x - 3 = 0$ …………………………….(1)
Now we put the option one by one and check
We now check \[( - 4,4)\] satisfy the above equation (1) or not
Therefore L.H.S. $2 \times 4 + ( - 4) - 3$
$ = 8 - 4 - 3$
$ = 1 \ne R.H.S.$
$\therefore $ Option (1) is incorrect .
Now check $(1, - 2)$ satisfy the above equation (1) or not
L.H.S. $2 \times ( - 2) + 1 - 3$
$ = - 4 + 1 - 3$
$ = - 6$ $ \ne R.H.S.$
$\therefore $ Option (2) is incorrect .
Now check $\left( {\dfrac{9}{4},\dfrac{3}{8}} \right)$ satisfy the above equation (1) or not
L.H.S. $2 \times \dfrac{3}{8} + \dfrac{9}{4} - 3$
$ = \dfrac{3}{4} + \dfrac{9}{4} - 3$
$ = \dfrac{{3 + 9 - 3 \times 4}}{4}$
$ = \dfrac{{12 - 12}}{4}$
$ = 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 $Q$ .
> Only for multiple choice questions we use the above method.