Question

Slope of normal to the curve \[y = {x^2} - \dfrac{1}{{{x^2}}}\] at \[\left( { - 1,0} \right)\] is
A.\[\dfrac{1}{4}\]
B.\[ - \dfrac{1}{4}\]
C.\[4\]
D.\[ - 4\]

Answer 1

Hint: Slope of a curve is calculated by differentiating it with respect to the given variable, here it is given to be \[x\]. But this does not give us the answer to the question. On differentiating it, we get the slope of tangent. For finally calculating it for the slope of the normal, we apply the formula which converts the slope of tangent to the slope of normal.

Formula Used:
We will have to differentiate the equation of the curve, so we are going to need to apply the differentiation formula, which is:
\[\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}{\rm{ }}\]…(i)
Then, we will have to convert the slope of tangent to the slope of normal and the formula for the same is:
Let the slope of tangent calculated after differentiation be \[{m_t}\], then the slope of normal can be represented by:
\[{m_n} = - \dfrac{1}{{{m_t}}}\] …(ii)

Complete step-by-step answer:
In solving questions like these, it is always advised to bring all the variables in the denominator to the numerator, and this is achieved by reversing the powers of the variables which are to be brought to the numerator. The formula for the same is:
\[\dfrac{1}{{{a^n}}} = {a^{ - n}}\]
So first, we have to bring the \[{x^2}\] in the denominator of \[\dfrac{1}{{{x^2}}}\]to its numerator. So, it will become:
$\Rightarrow$ \[\dfrac{1}{{{x^2}}} = {x^{ - 2}}\]
So, the complete equation of the slope becomes
$\Rightarrow$ \[y = {x^2} - {x^{ - 2}}\]
Now applying the formula from (i) and differentiating, we have
$\Rightarrow$ \[\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {{x^2} - {x^{ - 2}}} \right)}}{{dx}}\]
Separating the two terms so as to evaluate them separately, we get
$\Rightarrow$ \[\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {{x^2}} \right)}}{{dx}} - \dfrac{{d\left( {{x^{ - 2}}} \right)}}{{dx}}\]
$\Rightarrow$ \[\dfrac{{dy}}{{dx}} = 2{x^{2 - 1}} - \left( { - 2{x^{ - 2 - 1}}} \right)\]
Now, applying the formula from (i),
$\Rightarrow$ \[\dfrac{{dy}}{{dx}} = 2x + 2{x^{ - 3}}\]
Now, it has been given in the question that we have to evaluate the slope at \[\left( { - 1,0} \right)\], but since we only have a single variable (x), we will only have to consider the \[x\], so
$\Rightarrow$ \[{\left. {\dfrac{{dy}}{{dx}}} \right|_{x = - 1}} = 2\left( { - 1} \right) + 2{\left( { - 1} \right)^{ - 3}} = - 2 + \left( { - 2} \right) = - 2 - 2 = - 4\]
Hence, the slope of the tangent is \[ - 4\], so the slope of normal is (by applying the formula from (ii)):
$\Rightarrow$ \[{m_{normal}} = - \dfrac{1}{{\left( { - 4} \right)}} = \dfrac{1}{4}\]
Hence, the correct option is A \[\dfrac{1}{4}\].

Note: For calculating the slope, we first differentiate the given expression, put in the given values and we will have the slope of tangent and for calculating the slope of normal, we take the reciprocal of the value and invert its sign. This is going to give us the slope of the normal.