Angle subtended by a chord of a parabola to its vertex

Question

The equations of the two chords of the parabola y^2=4ax are to be found such that the pass through the point (-6a,0) and subtends an angle of 45 degree at the vertex of the parabolas. Tried it in many ways including using parametric equations but could not get the two equations

(note. Let the chord intersect the parabola at points p and q. And let the vertex be point r then prq..=45 degree)

Answer 1

Hint:

Note that the vertex of the parabola $y^2=4ax$ is the point $(0,0)$. The equation of the chord in slope-intercept form is given by $$y = kx + b$$ where $k$ is the slope of the line which is given to be equal to $\tan 45^\circ = 1$. Now, use the fact that the chord also passes through $(-6a,0)$. Can you take it from here?

Answer 2

The vertex of the parabola is at the origin and its axis is the $x$-axis. A line through the vertex with slope $m\ne0$ intersects the parabola at $P=\left({4a\over m^2},{4a\over m}\right)$. The line with a slope of $-\frac1m$ is perpendicular to it, so the 45-degree lines have slopes given by the two angle bisectors ${m-1\over m+1}$ and $-{m+1\over m-1}$ (The latter is vertical for $m=1$, but we know that can’t be one of the solutions because that’s tangent to the parabola.) Taking either of these lines, let $Q$ be the other intersection of this line with the parabola. The point $R=(-6a,0)$ lies on the line $\overline{PQ}$ iff the matrix $$\begin{bmatrix}x_P&y_P&1\\x_Q&y_Q&1\\x_R&y_R&1\end{bmatrix}$$ has rank 2. Row-reduction (or computing the determinant) will give you a quartic equation in $m$, but factoring it isn’t too hard. You should end up with two real roots.

Answer 3

$y^2=4ax$ is a parabola.

Let chord PQ passes from $(-6a,0)$ and subtends angle 45 degree at the vertex i.e. R

Now,

let PQ be $y=m(x+6a)$ ---{1}

and now using method of homogenization.

We get, $y^2=4ax(y-mx/6am)$ by solving it we get,

$3m(y/x)^2-2(y/x)+2m=0$ --{2}

Now this is quadratic equation which have roots $m_1$ and $m_2$. And these are the slopes of two other chords which are PR & QR. And as given in question PR & QR have 45 degree angle between them. So,

$tanA=|m_1-m_2/1+m_1m_2|$ ---{3}

and A=45 degree(given in Ques.)and sum of roots $m_1+m_2=2/3m$ and product of roots $m_1m_2=2/3$ from{1}.

So, by solving{2}

We get $m=2/3$ and $m=-2/3$

Therefore putting value of m in{1}

We get equations of PR & QR

$y=2/3(x+6a)$ and $y=-2/3(x+6a)$