At what angle does the line $x=a$ cut the curve $x^2=4y$?

Question

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.

Answer 1

Instead of thinking in terms of slope formulas, it's probably easier to use tangent vectors at the point of intersection, since you can then leverage the dot product formula to find the angle between these two vectors.

The tangent vector for the vertical line $x = a$ can be chosen to be pointing vertically upwards, $⟨0, 1⟩$, or vertically downwards, $⟨0, -1⟩$.

The tangent vector for the parabola $x^2 = 4y$ can be found by parametrizing this curve in terms of $t$: $$\overrightarrow{r}(t) = ⟨t, t^2/4⟩,$$ so that its tangent vector will then be $$\overrightarrow{r'}(t) = ⟨1, t/2⟩,$$ and from there you can plug in $t = a$ at the intersection point to find the specific tangent vector.

Answer 2

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,$$\theta=\arctan\left(\frac2a\right)=\operatorname{arccot}\left(\frac a2\right).$$