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The lengths of 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
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The correct answer is:
$B, A, D, C$
Given circle is $y=x^2+x-1$ and point
$\begin{aligned} & \left(x_1, y_1\right)=(1,1) \\ & \therefore \frac{d y}{d x}=2 x+1\end{aligned}$
At $(1,1), \quad \frac{d y}{d x}=3=m$
Now, length of tangent
$A=\left|\frac{y_1 \sqrt{1+m^2}}{m}\right|=\left|\frac{1 \sqrt{1+9}}{3}\right|=\frac{\sqrt{10}}{3}$
length of subtangent $B=\left|\frac{y_1}{m}\right|=\frac{1}{3}$
length of normal
$C=\left|y_1 \sqrt{1+m^2}\right|=|1 \sqrt{1+9}|=\sqrt{10}$
and length of subnormal $D=\left|y_1 m\right|=3$ Now, increasing order is $B, A, D, C$.
$\begin{aligned} & \left(x_1, y_1\right)=(1,1) \\ & \therefore \frac{d y}{d x}=2 x+1\end{aligned}$
At $(1,1), \quad \frac{d y}{d x}=3=m$
Now, length of tangent
$A=\left|\frac{y_1 \sqrt{1+m^2}}{m}\right|=\left|\frac{1 \sqrt{1+9}}{3}\right|=\frac{\sqrt{10}}{3}$
length of subtangent $B=\left|\frac{y_1}{m}\right|=\frac{1}{3}$
length of normal
$C=\left|y_1 \sqrt{1+m^2}\right|=|1 \sqrt{1+9}|=\sqrt{10}$
and length of subnormal $D=\left|y_1 m\right|=3$ Now, increasing order is $B, A, D, C$.
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