Question

The length of the arc of the parabola ${x^2} = 4ay$ measured from the vertex to one extremity of the latus rectum is:
(A) $a\left[ {\sqrt 2 + \log \left( {1 - \sqrt 2 } \right)} \right]$
(B) $a\left[ {\sqrt 2 + \log \left( {1 + \sqrt 2 } \right)} \right]$
(C) $a\left[ {\sqrt 2 - \log \left( {1 + \sqrt 2 } \right)} \right]$
(D) $a\left[ {\sqrt 2 - \log \left( {1 - \sqrt 2 } \right)} \right]$

Answer 1

In the given question, we are required to find the length of the arc of the parabola ${x^2} = 4ay$ measured from the vertex to one extremity of the latus rectum. The length of arc of the parabola from vertex to one of the extremities can be calculated by firstly finding the derivative of y and then applying the Pythagoras theorem find the length of arc by integrating the same from $0$ to $2a$.

Complete step by step answer:
Let A be the vertex and L be an extremity of the latus rectum so that at A$\left( {0,0} \right)$ and L$\left( {2a,a} \right)$.Now, given
$y = \left( {\dfrac{{{x^2}}}{{4a}}} \right)$
So, on differentiating both sides of the equation of parabola given to us in the question, we get,
$\dfrac{{dy}}{{dx}} = \left( {\dfrac{1}{{4a}}} \right)\left( {2x} \right)$
$\Rightarrow \dfrac{{dy}}{{dx}} = \left( {\dfrac{x}{{2a}}} \right)$
Hence, the length of the arc of the parabola ${x^2} = 4ay$ measured from the vertex to one extremity of the latus rectum can now be found using the Pythagoras theorem.
Length of arc $= \int_0^{2a} {\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} } dx$

Now, substituting the value of $\dfrac{{dy}}{{dx}}$ from the previously obtained equation, we get,
$\Rightarrow \int_0^{2a} {\sqrt {1 + {{\left( {\dfrac{x}{{2a}}} \right)}^2}} } dx$
$\Rightarrow \left( {\dfrac{1}{{2a}}} \right)\int_0^{2a} {\sqrt {{{\left( {2a} \right)}^2} + {x^2}} } dx$
Now, the integral that we need to calculate in order to find the final answer is a special form integral $\int {\sqrt {{x^2} + {a^2}} dx = \dfrac{x}{2}} \sqrt {{x^2} + {a^2}} + \dfrac{{{a^2}}}{2}\log \left( {x + \sqrt {{x^2} + {a^2}} } \right)$. So, we simplify the integral using the result of this special integral.
$\Rightarrow \left( {\dfrac{1}{{2a}}} \right)\int_0^{2a} {\sqrt {{{\left( {2a} \right)}^2} + {x^2}} } dx = \left( {\dfrac{1}{{2a}}} \right){\left[ {\dfrac{x}{2}\sqrt {{x^2} + 4{a^2}} + \dfrac{{4{a^2}}}{2}\log \left( {x + \sqrt {{x^2} + 4{a^2}} } \right)} \right]^{2a}}_0$

Now, putting in the upper limit and lower limit of the definite integral and computing the length of arc,
\Rightarrow \left( {\dfrac{1}{{2a}}} \right)\left\\{ {\left[ {\dfrac{{2a}}{2}\sqrt {{{\left( {2a} \right)}^2} + 4{a^2}} + \dfrac{{4{a^2}}}{2}\log \left( {2a + \sqrt {{{\left( {2a} \right)}^2} + 4{a^2}} } \right)} \right] - \left[ {\dfrac{0}{2}\sqrt {{0^2} + 4{a^2}} + \dfrac{{4{a^2}}}{2}\log \left( {0 + \sqrt {{0^2} + 4{a^2}} } \right)} \right]} \right\\}
\Rightarrow \left( {\dfrac{1}{{2a}}} \right)\left\\{ {\left[ {a\sqrt {8{a^2}} + 2{a^2}\log \left( {2a + \sqrt {8{a^2}} } \right)} \right] - \left[ {0 + 2{a^2}\log \left( {\sqrt {4{a^2}} } \right)} \right]} \right\\}

Simplifying further, we get,
\left( {\dfrac{1}{{2a}}} \right)\left\\{ {\left[ {2\sqrt 2 {a^2} + 2{a^2}\log \left( {2a + 2\sqrt 2 a} \right)} \right] - \left[ {0 + 2{a^2}\log \left( {2a} \right)} \right]} \right\\}
\Rightarrow \left( {\dfrac{1}{{2a}}} \right)\left\\{ {2\sqrt 2 {a^2} + 2{a^2}\log \left( {2a + 2\sqrt 2 a} \right) - 2{a^2}\log \left( {2a} \right)} \right\\}
\Rightarrow \left( {\dfrac{1}{{2a}}} \right)\left\\{ {2\sqrt 2 {a^2} + 2{a^2}\left[ {\log \left( {2a + 2\sqrt 2 a} \right) - \log \left( {2a} \right)} \right]} \right\\}
\Rightarrow \left( {\dfrac{1}{{2a}}} \right)\left\\{ {2\sqrt 2 {a^2} + 2{a^2}\left[ {\log \left( {1 + \sqrt 2 } \right)} \right]} \right\\}
Cancelling $2a$ in numerator and denominator, we get,
$\therefore a\left( {\sqrt 2 + \log \left( {1 + \sqrt 2 } \right)} \right)$

Hence, the length of the arc of the parabola ${x^2} = 4ay$ measured from the vertex to one extremity of the latus rectum is: $a\left( {\sqrt 2 + \log \left( {1 + \sqrt 2 } \right)} \right)$.

Note: The above solution represents the standard method for solving such questions that require us to find the length of an arc or length of a curve given the equation of the curve. These types of questions require knowledge of applications of derivatives and a strong grip on calculus and analytical skills.