Let [PQ] be a focal chord of the parabola y^2=36 x of length 100 , making an acute angle w…

Question

Let $$\mathrm{PQ}$$ be a focal chord of the parabola $y^{2}=36 x$ of length 100 , making an acute angle with the positive $x$-axis. Let the ordinate of $$\mathrm{P}$$ be positive and $$\mathrm{M}$$ be the point on the line segment PQ such that PM : MQ = 3 : 1. Then which of the following points does NOT lie on the line passing through M and perpendicular to the line $$\mathrm{PQ}$$?

Accepted Answer

The given parabola is of the form $y^2 = 4ax$. Here, 4a = 36, which means a = 9.

Length of focal chord at $(t)=a\left(t+\frac{1}{t}\right)^2=100$

Where $a=9$

$$
\begin{gathered}
t+\frac{1}{t}= \pm \frac{10}{3} \\
\therefore \quad t=3, \frac{1}{3},-3, \frac{-1}{3}
\end{gathered}
$$

Since ordinate of $P$ is $+\mathrm{ve}$

$$
\therefore t=3 \text { or } \frac{1}{3}
$$

For each value of t, we can find the coordinates of the points P and Q on the parabola using the parametric form of the parabola $y^2=4ax$, where a is 9. The parametric form is given by $x=at^2$ and $y=2at$.

Let's consider $t=3$ first.

For $t=3$, the coordinates of P are given by $P(at^2, 2at) = P(81, 54)$.

Since $t_1t_2=-1$ for a focal chord, the parameter value for Q is $t_2=-1/t_1=-1/3$. So the coordinates of Q are given by $Q(at_2^2, 2at_2) = Q(1, -6)$.

Now, let's find the coordinates of the point M that divides the line segment PQ in the ratio 3:1.

The x-coordinate of M is given by $\frac{3Q_x+P_x}{4} = \frac{3*1+81}{4} = 21$.

The y-coordinate of M is given by $\frac{3Q_y+P_y}{4} = \frac{3*(-6)+54}{4} = 9$.

So, the coordinates of M are M$(21, 9)$.

We can repeat these steps for $t=1/3$ to find the other set of points P, Q, and M. However, since the ordinate of P is positive and PQ makes an acute angle with the positive x-axis, $t=3$ is the appropriate choice in this context.

The slope of the line PQ is $m_{PQ} = \frac{-6 - 54}{1 - 81} = \frac{-60}{-80} = \frac{3}{4}$. The slope of the line perpendicular to PQ is $m_{\perp PQ} = -1/m_{PQ} = -\frac{4}{3}$. Thus, the equation of the line through M and perpendicular to PQ is $(y - 9) = -\frac{4}{3}(x - 21)$, or $4x + 3y = 111$.

Now, we check which of the given points do NOT lie on this line.

Option A : $(6, 29)$
Substitute these values into the equation :

$4\times6 + 3\times29 = 111?$

$24 + 87 = 111?$

$111 = 111$

So, option A does lie on the line.

Option B : $(-3, 43)$
Substitute these values into the equation :

$4\times(-3) + 3\times43 = 111?$

$-12 + 129 = 111?$

$117 = 111$

So, option B does NOT lie on the line.

Option C : $(3, 33)$
Substitute these values into the equation :

$4\times3 + 3\times33 = 111?$

$12 + 99 = 111?$

$111 = 111$

So, option C does lie on the line.

Option D : $(-6, 45)$
Substitute these values into the equation :

$4\times(-6) + 3\times45 = 111?$

$-24 + 135 = 111?$

$111 = 111$

So, option D does lie on the line.

Therefore, $(-3, 43)$ which does not lie on the line passing through M and perpendicular to the line PQ.

Solutions · 1