Question

The length of a tangent, subtangent, normal and subnormal for the curve $y={{x}^{2}}+x-1$ at (1,1) are A, B, C, and D respectively, then their increasing order is.
(a) B, D, A, C
(b) B, A, C, D
(c) A, B, C, D
(d) B, A, D, C

Answer 1

Hint: For solving this question first we will see the formulas for the length of a tangent, subtangent, normal, subnormal. After that, we will differentiate it with respect to $x$ and calculate the value of $\dfrac{dy}{dx}$. Then, we will directly find the length of the subtangent from its formula.

Complete step-by-step solution
Given:
We have to find the increasing order of length of a tangent, subtangent, normal, and subnormal for the curve $y={{x}^{2}}+x-1$ at the point (1,1). And it is given that A is the length of a tangent, B is the length of a subtangent, C is the length of normal and D is the length of subnormal.
Now, before we proceed we should know the following four formulas:
1. Length of tangent for any curve $y=f\left( x \right)$ at a point $\left( {{x}_{1}},{{y}_{1}} \right)$ on the curve is equal to ${{\left| y\sqrt{1+{{\left( \dfrac{dx}{dy} \right)}^{2}}} \right|}_{\left( {{x}_{1}},{{y}_{1}} \right)}}$ .
2. Length of subtangent for any curve $y=f\left( x \right)$ at a point $\left( {{x}_{1}},{{y}_{1}} \right)$ on the curve is equal to ${{\left| y\dfrac{dx}{dy} \right|}_{\left( {{x}_{1}},{{y}_{1}} \right)}}$ .
3. Length of normal for any curve $y=f\left( x \right)$ at a point $\left( {{x}_{1}},{{y}_{1}} \right)$ on the curve is equal to ${{\left| y\sqrt{1+{{\left( \dfrac{dy}{dx} \right)}^{2}}} \right|}_{\left( {{x}_{1}},{{y}_{1}} \right)}}$ .
4. Length of subnormal for any curve $y=f\left( x \right)$ at a point $\left( {{x}_{1}},{{y}_{1}} \right)$ on the curve is equal to ${{\left| y\dfrac{dy}{dx} \right|}_{\left( {{x}_{1}},{{y}_{1}} \right)}}$ .
Now, first, we will find the value of $\dfrac{dy}{dx}$ at point (1,1) for the function $y={{x}^{2}}+x-1$ . Then,
$\begin{align}
  & y={{x}^{2}}+x-1 \\
 & \Rightarrow \dfrac{dy}{dx}=2x+1 \\
 & \Rightarrow {{\left[ \dfrac{dy}{dx} \right]}_{\left( 1,1 \right)}}=2+1 \\
 & \Rightarrow {{\left[ \dfrac{dy}{dx} \right]}_{\left( 1,1 \right)}}=3 \\
\end{align}$
Now, using the formulas for the length of tangent, subtangent, normal, subnormal to find the value of A, B, C, D. Then,
\[\begin{align}
  & A={{\left| y\sqrt{1+{{\left( \dfrac{dx}{dy} \right)}^{2}}} \right|}_{\left( 1,1 \right)}}={{\left| y\sqrt{1+\dfrac{1}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}} \right|}_{\left( 1,1 \right)}} \\
 & \Rightarrow A=\sqrt{1+\dfrac{1}{9}}=\sqrt{\dfrac{10}{9}} \\
 & \Rightarrow A=\dfrac{\sqrt{10}}{3} \\
 & B={{\left| y\dfrac{dx}{dy} \right|}_{\left( 1,1 \right)}}={{\left| y\times \dfrac{1}{\dfrac{dy}{dx}} \right|}_{\left( 1,1 \right)}} \\
 & \Rightarrow B=\dfrac{1}{3} \\
 & C={{\left| y\sqrt{1+{{\left( \dfrac{dy}{dx} \right)}^{2}}} \right|}_{\left( 1,1 \right)}}=\left| 1\times \sqrt{1+9} \right| \\
 & \Rightarrow C=\sqrt{10} \\
 & D={{\left| y\dfrac{dy}{dx} \right|}_{\left( 1,1 \right)}} \\
 & \Rightarrow D=3 \\
\end{align}\]
Now, from the above result, we conclude that $B< A< D< C$ .
Hence, (d) is the correct option.

Note: Here, the student should apply the formula for the length of a tangent, subtangent, normal, and subnormal directly and proceed in a stepwise manner. But we should be careful while writing their formulas as they might seem to be similar. Moreover, we should substitute correct values while calculating to get the correct answer.