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If tangent to the curve $x=a t^{2}, y=2 a t$ is perpendicular to $X$-axis, then its point of contact is
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Verified Answer
The correct answer is:
$(0,0)$
We have,
$$
x=a t^{2}, y=2 a t
$$
Now, $\quad \frac{d x}{d t}=2 a t, \frac{d y}{d t}=2 a$
$$
\therefore \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{2 a}{2 a t}=\frac{1}{t}
$$
Since, tangent is perpendicular to $X$-axis, then
$\frac{d y}{d x}=\frac{1}{0}$
$\Rightarrow \quad \frac{1}{t}=\frac{1}{0} \Rightarrow t=0$
$\therefore \quad x=a t^{2}=0$ and $y=2 a t=0$
So, required point is $(0,0)$.
$$
x=a t^{2}, y=2 a t
$$
Now, $\quad \frac{d x}{d t}=2 a t, \frac{d y}{d t}=2 a$
$$
\therefore \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{2 a}{2 a t}=\frac{1}{t}
$$
Since, tangent is perpendicular to $X$-axis, then
$\frac{d y}{d x}=\frac{1}{0}$
$\Rightarrow \quad \frac{1}{t}=\frac{1}{0} \Rightarrow t=0$
$\therefore \quad x=a t^{2}=0$ and $y=2 a t=0$
So, required point is $(0,0)$.
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