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Angle Subtended by an Arc of a Circle on the Circle and at the Center

The chord A...

Question

The chord $A C$ of the parabola $y^{2} = 4 a x$ subtends angles of $90^{\circ}$ at points $B$ and $D$ on the parabola. If $A, B, C$ and $D$ are respresented by the points $t_{1}$ , $t_{2}$ , $t_{3}$ and $t_{4}$ respectively, then

A

∣ ∣ ∣ \frac{t_{2} + t_{4}}{t_{1} + t_{3}} ∣ ∣ ∣ = 2

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B

| t_{2} t_{4} - t_{1} t_{3} | = 4

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C

| t_{1} - t_{3} | > 4

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D

The y-coordinate of the mid point of the points of intersection of the tangents at

A, C

and

B, D

is

0

.

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Solution

The correct options are
A $| t_{2} t_{4} - t_{1} t_{3} | = 4$
C $| t_{1} - t_{3} | > 4$
D The y-coordinate of the mid point of the points of intersection of the tangents at $A, C$ and $B, D$ is $0$ .
Slope of $A B$ is $\frac{2}{t_{1} + t_{2}}$
Similarly, slope of $B C$ is $\frac{2}{t_{2} + t_{3}}$
Since $A B$ and $B C$ are perpendicular, we get,
We have $(t_{2} + t_{3}) (t_{2} + t_{1}) = - 4$
$\Rightarrow {t_{2}}^{2} + (t_{1} + t_{3}) t_{2} + t_{1} t_{3} + 4 = 0 . . . . . . . . . (1)$
Since there are two points $B$ and $D$ for which the condition is satisfied,
the above equation has two real roots.
$\Rightarrow D > 0$

$\Rightarrow (t_{1} + t_{3})^{2} - 4.1 (t_{1} t_{3} + 4) > 0$
$\Rightarrow (t_{1} - t_{3})^{2} > 16$
$\Rightarrow | t_{1} - t_{3} | > 4$
Also, $t_{2} + t_{4} = - (t_{1} + t_{3})$ and $t_{2} t_{4} = t_{1} t_{3} + 4$
$\Rightarrow t_{1} + t_{2} + t_{3} + t_{4} = 0$ and $t_{2} t_{4} - t_{1} t_{3} = 4$
(because $t_{2}$ and $t_{4}$ are the roots of equation (1))
Also let $y_{1}$ and $y_{2}$ are y-coordinates of intersection points of the
tangents at (A,C) and (B, D), then
$y_{1} = a (t_{1} + t_{3}), y_{2} = a (t_{2} + t_{4})$
$\Rightarrow$ 'y' coordinate of the midpoint is $\frac{y_{1} + y_{2}}{2} = \frac{a (t_{1} + t_{2} + t_{3} + t_{4})}{2} = 0$

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