Question

If the tangent at $\left( {1,1} \right)$ on ${y^2} = x{\left( {2 - x} \right)^2}$ meets the curve again at $P$, then find the coordinates of $P$?

Answer 1

Here in this question, we have to find the point P where the curve meets, to solve this first we have to find the slope at the given point by differentiating the equation of curve i.e., find $m = {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {1,1} \right)}}$ and later substitute a value of slope in a equation of tangent $y - {y_0} = m\left( {x - {x_0}} \right)$ where $\left( {{x_0},{y_0}} \right)$ is the point where tangent line passes through and further simplify by method of factorization we get the required solution.

Complete step by step solution:
The tangent line to a curve at a given point is a straight line that just "touches" the curve at that point.
Considering the given equation of curve
${y^2} = x{\left( {2 - x} \right)^2}$---------(1)
Expand the RHS using a algebraic identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, then
$\Rightarrow \,\,{y^2} = x\left( {{2^2} + {x^2} - 2 \cdot 2 \cdot x} \right)$
$\Rightarrow \,\,{y^2} = x\left( {4 + {x^2} - 4x} \right)$
Multiplying $x$ into the parenthesis on RHS, we get.
$\Rightarrow \,\,{y^2} = 4x + {x^3} - 4{x^2}$------(2)
The tangent line passes at the point $\left( {1,1} \right)$
Now, differentiate the equation (2) with respect to x , and we get a slope of the line.
$\Rightarrow \,\,\dfrac{d}{{dx}}\left( {{y^2}} \right) = \dfrac{d}{{dx}}\left( {4x + {x^3} - 4{x^2}} \right)$
$\Rightarrow \,\,\dfrac{d}{{dx}}\left( {{y^2}} \right) = 4\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( {{x^3}} \right) - 4\dfrac{d}{{dx}}\left( {{x^2}} \right)$
Differentiate using a formula $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$, then
$\Rightarrow \,\,2y\dfrac{{dy}}{{dx}} = 4\left( 1 \right) + 3{x^2} - 4\left( {2x} \right)$
$\Rightarrow \,\,2y\dfrac{{dy}}{{dx}} = 4 + 3{x^2} - 8x$
Divide both side by 2y, then
$\Rightarrow \,\,\dfrac{{dy}}{{dx}} = \dfrac{{4 + 3{x^2} - 8x}}{{2y}}$
Now find the slope $m = \dfrac{{dy}}{{dx}}$ at the point $\left( {1,1} \right)$, then
$\Rightarrow \,\,m = {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {1,1} \right)}} = \dfrac{{4 + 3{{\left( 1 \right)}^2} - 8\left( 1 \right)}}{{2\left( 1 \right)}}$
$\Rightarrow \,\,m = {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {1,1} \right)}} = \dfrac{{4 + 3 - 8}}{2}$
$\Rightarrow \,\,m = {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {1,1} \right)}} = - \dfrac{1}{2}$
Now, put equation of tangent lies at the point $\left( {1,1} \right)$ is:
The equation of tangent is $y - {y_0} = m\left( {x - {x_0}} \right)$
Here ${y_0} = 1$, ${x_0} = 1$ and $m = - \dfrac{1}{2}$, on substituting we get
$\Rightarrow \,\,y - 1 = \left( { - \dfrac{1}{2}} \right)\left( {x - 1} \right)$
Multiply both side by 2, then
$\Rightarrow \,\,2\left( {y - 1} \right) = - 1\left( {x - 1} \right)$
On simplification, we get
$\Rightarrow \,\,2y - 2 = - x + 1$
Take all term in RHS to LHS, then
$\Rightarrow \,\,2y - 2 + x - 1 = 0$
$\Rightarrow \,\,x + 2y - 3 = 0$--------(3)
Add 3 and subtract 2y on both sides, then
$\Rightarrow \,\,x = 3 - 2y$--------(4)
Now, substitute equation (4) in the equation of cure i.e., equation (1), we get.
$\Rightarrow \,\,{y^2} = \left( {3 - 2y} \right){\left( {2 - \left( {3 - 2y} \right)} \right)^2}$
$\Rightarrow \,\,{y^2} = \left( {3 - 2y} \right){\left( {2 - 3 + 2y} \right)^2}$
$\Rightarrow \,\,{y^2} = \left( {3 - 2y} \right){\left( { - 1 + 2y} \right)^2}$
Or
$\Rightarrow \,\,{y^2} = \left( {3 - 2y} \right){\left( {2y - 1} \right)^2}$
Expand RHS using a algebraic identity: ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, then
$\Rightarrow \,\,{y^2} = \left( {3 - 2y} \right)\left( {{{\left( {2y} \right)}^2} + {1^2} - 2\left( {2y} \right)1} \right)$
$\Rightarrow \,\,{y^2} = \left( {3 - 2y} \right)\left( {4{y^2} + 1 - 4y} \right)$
$\Rightarrow \,\,{y^2} = 3\left( {4{y^2} + 1 - 4y} \right) - 2y\left( {4{y^2} + 1 - 4y} \right)$
On simplification, we have
$\Rightarrow \,\,{y^2} = 12{y^2} + 3 - 12y - 8{y^3} - 2y + 8{y^2}$
Take RHS in to LHS, then
$\Rightarrow \,\,{y^2} - 12{y^2} - 3 + 12y + 8{y^3} + 2y - 8{y^2} = 0$
$\Rightarrow \,\,8{y^3} - 19{y^2} + 14y - 3 = 0$
It can be re-written as
$\Rightarrow \,\,8{y^3} - 8{y^2} - 11{y^2} + 11y + 3y - 3 = 0$
Take out the GCD, then
$\Rightarrow \,\,8{y^2}\left( {y - 1} \right) - 11y\left( {y - 1} \right) + 3\left( {y - 1} \right) = 0$
Take $\left( {y - 1} \right)$ as common, then
$\Rightarrow \,\,\left( {y - 1} \right)\left( {8{y^2} - 11y + 3} \right) = 0$-----(5)
Now, factorize
$\Rightarrow \,\,8{y^2} - 11y + 3$
$\Rightarrow \,\,8{y^2} - 8y - 3y + 3$
$\Rightarrow \,\,8y\left( {y - 1} \right) - 3\left( {y - 1} \right)$
$\Rightarrow \,\,\left( {y - 1} \right)\left( {8y - 3} \right)$
Then, equation (5) becomes
$\Rightarrow \,\,\left( {y - 1} \right)\left( {y - 1} \right)\left( {8y - 3} \right) = 0$
Then,
$\Rightarrow \,\,y = 1,\,1,\,\dfrac{3}{8}$
Take the value $y = \dfrac{3}{8}$
Substituting $y$ value in equation (4), then
$\Rightarrow \,\,x = 3 - 2\left( {\dfrac{3}{8}} \right)$
$\Rightarrow \,\,x = 3 - \dfrac{3}{4}$
Take 4 as LCM, then
$\Rightarrow \,\,x = \dfrac{{12 - 3}}{4}$
$\Rightarrow \,\,x = \dfrac{9}{4}$
Therefore, the coordinates of point $P$ is $\left( {x,y} \right) = \left( {\dfrac{9}{4},\dfrac{3}{8}} \right)$.

Note:
The concept of the equation of tangent comes under the concept of application of derivatives. Here the major part is differentiation must know the standard differentiation formulas and we should know about the general equation of a line tangent. Hence these types of problems are solved by the above procedure and we can also find the equation of tangent using the above procedure.