The line $x=a$ cuts $x^2=4y$ at an angle of :
The options mentioned in the question sheet are :
a) $\arctan\left(\frac{a}{4}\right)$
b) $\operatorname{arccot}\left(\frac{a}{2}\right)$
c) $\left(\frac{1}{2}\right)\cdot \arctan(a)$
d) $\arctan\left(\frac{a}{2}\right)$
I tried to find $\frac{{dy}}{{dx}}$
at $x=a$ to obtain the slope of the tangent to the curve $x^2=4y$ at the point $x=a$.
$\frac{{dy}}{{dx}}$ at $x=a$ came out to be $\frac{a}{2}
$. After this, how would you proceed further to find the angle at which the line cuts the curve?
My concept for this would be to get the slope of curve and the slope of the line and then use:
$\tan(\theta) = \pm \frac{{m_1 – m_2}}{{1 + m_1 \cdot m_2}}$
where $m_1$= slope of curve i.e. $x^2=4y$
$m_2$= slope of the line i.e. $x=a$
But the line $x=a$ will have undefined slope, so I am stuck.
Requesting you all to guide me through the concept and solution of this question.
Thank you.
Best Answer
The line $x=a$ cuts the curve $x^2=4y$ at the point $\left(a,\frac{a^2}4\right)$. The slope of the tangent line to that curve at that point is $\frac a2$. So, you are after the angle between a vertical line and a line whose slope is $\frac a2$. If $\theta$ is their angle, then $\theta=\frac\pi2-\arctan\left(\frac a2\right)$, and therefore\begin{align}\tan\theta&=\tan\left(\frac\pi2-\arctan\left(\frac a2\right)\right)\\&=\cot\left(\arctan\left(\frac a2\right)\right)\\&=\frac2a.\end{align}So,$$\theta=\arctan\left(\frac2a\right)=\operatorname{arccot}\left(\frac a2\right).$$