Question

What normal to the curve $y = {x^2}$ forms the shortest chord?

Answer 1

Here, in the given question, we need to find the equation of the shortest chord that is normal to the tangent. As we can see the given equation of the curve, $y = {x^2}$ is of parabola. In this given parabola equation the equation of the normal can be found by finding the slope of the normal and this normal is a chord so it will pass through two points in the parabola. Let points be $P$ and $Q$. We will find the coordinates of these two points. After this, we will find the distance between these two points using distance formula, and then we will differentiate it and find the minimum value (we know that for maximum or minimum value we always have first derivative equal to zero) because we need to find the shortest chord, from here we will get the value of coordinates. After this, we will substitute the values of coordinates in the general equation of the line to find the equation of chord.

Complete step by step answer:

The given equation of the curve is $y = {x^2}....\left( i \right)$
Differentiating $\left( i \right)$ with respect to $x$, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = 2x$
Any point on the curve $y = {x^2}$ is of the form $\left( {t,{t^2}} \right)$.
$\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {x = t} \right)}} = 2t$
Which is the slope of the tangent.

As we know tangent and normal are always perpendicular to each other. Therefore,
Slope of tangent × Slope of normal chord = $- 1$
So, the slope of the normal to $y = {x^2}$ at $P\left( {t,{t^2}} \right)$ is $\dfrac{{ - 1}}{{2t}}$.
As we know the general equation of line is $y - {y_1} = m\left( {x - {x_1}} \right)$.Therefore, the equation of the normal to $y = {x^2}$ at $P\left( {t,{t^2}} \right)$ is
$\Rightarrow y - {t^2} = \left( { - \dfrac{1}{{2t}}} \right)\left( {x - t} \right).....\left( {ii} \right)$
Suppose equation $\left( {ii} \right)$ meets the curve again at $Q\left( {{t_1},{t_1}^2} \right)$
$\Rightarrow {t_1}^2 - {t^2} = - \dfrac{1}{{2t}}\left( {{t_1} - t} \right)$
Using identity ${a^2} - {b^2}\left( {a - b} \right)\left( {a + b} \right)$, we get
$\Rightarrow \left( {{t_1} - t} \right)\left( {{t_1} + t} \right) = - \dfrac{1}{{2t}}\left( {{t_1} - t} \right)$
On cancelling common factors, we get
$\Rightarrow \left( {{t_1} + t} \right) = - \dfrac{1}{{2t}}$
$\Rightarrow {t_1} = - \dfrac{1}{{2t}} - t$

Now, we will find the length of the chord using the distance formula. Let L be the length of the chord.
$P{Q^2} = {\left( {{x_2} - {x_1}} \right)^2} + {\left( {{y_2} - {y_1}} \right)^2}$. Let point $Q\left( {{t_1},{t_1}^2} \right)$ be $\left( {{x_1},{y_1}} \right)$ and point $P\left( {t,{t^2}} \right)$ be $\left( {{x_2},{y_2}} \right)$.
$L = P{Q^2} = {\left( {t - {t_1}} \right)^2} + {\left( {{t^2} - {t_1}^2} \right)^2}$
Using identity ${a^2} - {b^2}\left( {a - b} \right)\left( {a + b} \right)$, we get
$\Rightarrow L = {\left( {t - {t_1}} \right)^2} + {\left( {t - {t_1}} \right)^2}{\left( {t + {t_1}} \right)^2}$
$\Rightarrow L = {\left( {t - {t_1}} \right)^2}\left[ {1 + {{\left( {t + {t_1}} \right)}^2}} \right]$
On substituting value of ${t_1}$ we get
$\Rightarrow L = {\left( {t + t + \dfrac{1}{{2t}}} \right)^2}\left[ {1 + {{\left( {t - t - \dfrac{1}{{2t}}} \right)}^2}} \right]$
On addition and subtraction of like terms, we get
$\Rightarrow L = {\left( {2t + \dfrac{1}{{2t}}} \right)^2}\left( {1 + \dfrac{1}{{4{t^2}}}} \right)$
$\Rightarrow L = 4{t^2}{\left( {1 + \dfrac{1}{{4{t^2}}}} \right)^3}$

Now, we will differentiate the length of the chord with respect to $t$ to get the value of $t$.
Derivative of product of two functions is given by the following product function rule:
$\dfrac{d}{{dx}}\left[ {f\left( x \right).g\left( x \right)} \right] = \dfrac{d}{{dx}}f\left( x \right).g\left( x \right) + f\left( x \right).\dfrac{d}{{dx}}g\left( x \right)$
On differentiating with respect to $t$, we get
$\Rightarrow \dfrac{{dL}}{{dt}} = 8t{\left( {1 + \dfrac{1}{{4{t^2}}}} \right)^3} + 4{t^2}.3{\left( {1 + \dfrac{1}{{4{t^2}}}} \right)^2}\left( { - \dfrac{2}{{4{t^3}}}} \right)$ ( As we know derivate of $\dfrac{1}{{{x^2}}} = \dfrac{{ - 2}}{{{x^3}}}$)
$\Rightarrow \dfrac{{dL}}{{dt}} = 8t{\left( {1 + \dfrac{1}{{4{t^2}}}} \right)^3} + 12{t^2}{\left( {1 + \dfrac{1}{{4{t^2}}}} \right)^2}\left( { - \dfrac{2}{{4{t^3}}}} \right)$

On taking $2{\left( {1 + \dfrac{1}{{4{t^2}}}} \right)^2}$ as common factor, we get
$\Rightarrow \dfrac{{dL}}{{dt}} = 2{\left( {1 + \dfrac{1}{{4{t^2}}}} \right)^2}\left[ {4t\left( {1 + \dfrac{1}{{4{t^2}}}} \right) - \dfrac{3}{t}} \right]$
Take LCM
$\Rightarrow \dfrac{{dL}}{{dt}} = 2{\left( {1 + \dfrac{1}{{4{t^2}}}} \right)^2}\left[ {4t\left( {\dfrac{{4{t^2} + 1}}{{4{t^2}}}} \right) - \dfrac{3}{t}} \right]$
$\Rightarrow \dfrac{{dL}}{{dt}} = 2{\left( {1 + \dfrac{1}{{4{t^2}}}} \right)^2}\left[ {\left( {\dfrac{{4{t^2} + 1}}{t}} \right) - \dfrac{3}{t}} \right]$
Take LCM
$\Rightarrow \dfrac{{dL}}{{dt}} = 2{\left( {1 + \dfrac{1}{{4{t^2}}}} \right)^2}\left[ {\dfrac{{4{t^2} + 1 - 3}}{t}} \right]$
$\Rightarrow \dfrac{{dL}}{{dt}} = 2{\left( {1 + \dfrac{1}{{4{t^2}}}} \right)^2}\left[ {\dfrac{{4{t^2} - 2}}{t}} \right]$

It can also be written as,
$\Rightarrow \dfrac{{dL}}{{dt}} = 2{\left( {1 + \dfrac{1}{{4{t^2}}}} \right)^2}\left( {4t - \dfrac{2}{t}} \right)$
Take $2$ as a common factor.
$\Rightarrow \dfrac{{dL}}{{dt}} = 4{\left( {1 + \dfrac{1}{{4{t^2}}}} \right)^2}\left( {2t - \dfrac{1}{t}} \right)$
For maxima and minima, we must have $\dfrac{{dL}}{{dt}} = 0$.
$4{\left( {1 + \dfrac{1}{{4{t^2}}}} \right)^2}\left( {2t - \dfrac{1}{t}} \right) = 0$
On dividing $4{\left( {1 + \dfrac{1}{{4{t^2}}}} \right)^2}$ on RHS we get,
$\Rightarrow 2t - \dfrac{1}{t} = 0$
On shifting $- \dfrac{1}{t}$ on RHS, we get
$\Rightarrow 2t = \dfrac{1}{t}$
$\Rightarrow {t^2} = \dfrac{1}{2}$

On taking square root on both sides, we get
$\Rightarrow t = \pm \dfrac{1}{{\sqrt 2 }}$
Next, $\dfrac{{{d^2}L}}{{d{t^2}}} = 8\left( {1 + \dfrac{1}{{4{t^2}}}} \right)\left( { - \dfrac{1}{{2{t^3}}}} \right)\left( {2t - \dfrac{1}{t}} \right) + 4{\left( {1 + \dfrac{1}{{4{t^2}}}} \right)^2}\left( {2 + \dfrac{1}{{{t^2}}}} \right)$
$\Rightarrow \dfrac{{{d^2}L}}{{d{t^2}}} = 8\left( {1 + \dfrac{1}{{4 \times \dfrac{1}{2}}}} \right)\left( { - \dfrac{1}{{2{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^3}}}} \right)\left( {2 \times \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{t}} \right) + 4{\left( {1 + \dfrac{1}{{4 \times \dfrac{1}{2}}}} \right)^2}\left( {2 + \dfrac{1}{{\dfrac{1}{2}}}} \right)$
As we can see $\dfrac{{{d^2}L}}{{d{t^2}}} > 0$ at $t = \pm \dfrac{1}{{\sqrt 2 }}$.Therefore, $L$ is minimum, when $t = \pm \dfrac{1}{{\sqrt 2 }}$. As we know we have points in the form of $t$. So now we will substitute the value of $t$ in the coordinates to get the coordinates in terms of the real number.

For $t = \left( {\dfrac{1}{{\sqrt 2 }}} \right)$, Point $P$ is $\left( {\dfrac{1}{{\sqrt 2 }},\dfrac{1}{2}} \right)$ and point $Q$ is $\left( { - \sqrt 2 ,2} \right)$. Now we have a passing point of the chord so we can find the equation of the chord. Let $P\left( {\dfrac{1}{{\sqrt 2 }},\dfrac{1}{2}} \right)$ be $\left( {{x_1},{y_1}} \right)$ and $Q\left( { - \sqrt 2 ,2} \right)$ be $\left( {{x_2},{y_2}} \right)$. Therefore,
Slope = $\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
Slope $= \dfrac{{2 - \dfrac{1}{2}}}{{ - \sqrt 2 - \dfrac{1}{{\sqrt 2 }}}}$
Take LCM of numerator and denominator
Slope $= \dfrac{{\dfrac{{4 - 1}}{2}}}{{\dfrac{{ - 2 - 1}}{{\sqrt 2 }}}} = \dfrac{{\dfrac{3}{2}}}{{\dfrac{{ - 3}}{{\sqrt 2 }}}}$
It can also be written as,
Slope $= \dfrac{3}{2} \times \dfrac{{\sqrt 2 }}{{ - 3}} = - \dfrac{1}{{\sqrt 2 }}$

We know the general equation of line is $y - {y_1} = m\left( {x - {x_1}} \right)$ using this we will find the equation of chord
$\Rightarrow y - \dfrac{1}{2} = \left( { - \dfrac{1}{{\sqrt 2 }}} \right)\left( {x - \dfrac{1}{{\sqrt 2 }}} \right)$
Take LCM on both the sides
$\Rightarrow \dfrac{{2y - 1}}{2} = \left( { - \dfrac{1}{{\sqrt 2 }}} \right)\left( {\dfrac{{\sqrt 2 x - 1}}{{\sqrt 2 }}} \right)$
$\Rightarrow \dfrac{{2y - 1}}{2} = - \left( {\dfrac{{\sqrt 2 \left( {x - 1} \right)}}{2}} \right)$
On cancelling common factors, we get
$\Rightarrow 2y - 1 = - \sqrt 2 \left( {x - 1} \right)$
$\Rightarrow 2y - 1 = - \sqrt 2 x + 1$

On shifting RHS terms to left side, we get
$\Rightarrow \sqrt 2 x + 2y - 1 - 1 = 0$
$\Rightarrow \sqrt 2 x + 2y - 2 = 0$
For $t = \left( { - \dfrac{1}{{\sqrt 2 }}} \right)$, Point $P$ is $\left( { - \dfrac{1}{{\sqrt 2 }},\dfrac{1}{2}} \right)$ and point $Q$ is $\left( {\sqrt 2 ,2} \right)$.
Slope $= \dfrac{{2 - \dfrac{1}{2}}}{{\sqrt 2 - \left( { - \dfrac{1}{{\sqrt 2 }}} \right)}}$
Take LCM of numerator and denominator
Slope $= \dfrac{{\dfrac{{4 - 1}}{2}}}{{\dfrac{{2 + 1}}{{\sqrt 2 }}}} = \dfrac{{\dfrac{3}{2}}}{{\dfrac{3}{{\sqrt 2 }}}}$
Slope $= \dfrac{3}{2} \times \dfrac{{\sqrt 2 }}{3} = \dfrac{1}{{\sqrt 2 }}$

We know the general equation of line is $y - {y_1} = m\left( {x - {x_1}} \right)$ using this we will find the equation of chord
$\Rightarrow y - \dfrac{1}{2} = \dfrac{1}{{\sqrt 2 }}\left( {x + \dfrac{1}{{\sqrt 2 }}} \right)$
Take LCM on both the sides
$\Rightarrow \dfrac{{2y - 1}}{2} = \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{{\sqrt 2 x + 1}}{{\sqrt 2 }}} \right)$
$\Rightarrow \dfrac{{2y - 1}}{2} = \dfrac{{\sqrt 2 x + 1}}{2}$
On canceling common factors, we get
$\Rightarrow 2y - 1 = \sqrt 2 x + 1$
Shift all the terms to one side
$\Rightarrow \sqrt 2 x - 2y + 1 + 1 = 0$
On adding terms, we get
$\therefore \sqrt 2 x - 2y + 2 = 0$

Hence, the equation of the shortest chord can be $\sqrt 2 x + 2y - 2 = 0$ or $\sqrt 2 x - 2y + 2 = 0$.

Note: Here, to find the derivative of length of the chord we used product rule because two functions were in the form of product but if we had to find the derivative of quotient of two functions we would have used quotient rule. Remember that for the minimum value, $f''\left( x \right)$ is greater than $0$ and for maximum value $f''\left( x \right)$ is less than $0$(Here $f''\left( x \right)$ is second derivative of any function).