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The number of points on the cure $y=2 t^2+3 t-5$ and $x=t^3-4 t^2-3 t$ such that the normals drawn at them on the curve are parallel to $\mathrm{X}$ - axis is
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Verified Answer
The correct answer is:
4
$\frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{4 t+3}{\left(3 t^2-8 t-3\right)}$
Hence slope of normal $=-\frac{1}{\left(\frac{d y}{d x}\right)}$ $=\frac{\left(3 t^2-8 t-3\right)}{(4 t+3)}$
Since slope of normal $=$ Slope of $x$-axis
$$
\begin{aligned}
& \Rightarrow \frac{3 t^2-8 t-3}{4 t+3}=0 \\
& \Rightarrow 3 t^2-8 t-3=0 \\
& \Rightarrow t=3 \text { and } t=-\frac{1}{3}
\end{aligned}
$$
Hence required number of points $=2$
Hence slope of normal $=-\frac{1}{\left(\frac{d y}{d x}\right)}$ $=\frac{\left(3 t^2-8 t-3\right)}{(4 t+3)}$
Since slope of normal $=$ Slope of $x$-axis
$$
\begin{aligned}
& \Rightarrow \frac{3 t^2-8 t-3}{4 t+3}=0 \\
& \Rightarrow 3 t^2-8 t-3=0 \\
& \Rightarrow t=3 \text { and } t=-\frac{1}{3}
\end{aligned}
$$
Hence required number of points $=2$
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