If $t_{1}$ and $t_{2}$ are the parameters of the endpoints of a focal chord for the parabola $y^{2} = 4 a x$ , then which one is correct.

$t_{1} t_{2} = 1$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{t_{1}}{t_{2}} = 1$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$t_{1} t_{2} = - 1$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$t_{1} + t_{2} = 1$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$t_{1} t_{2} = - 1$

Explanation for the correct option:
Let, that parabola $y^{2} = 4 a x$ .
Consider the endpoints of the parabola is $(a {t_{1}}^{2}, 2 a t_{1}) and (a {t_{2}}^{2}, 2 a t_{2})$ which are $P$ and $Q .$
The focus or the midpoint is going to be $(a, 0)$ .
Assume that $S .$
The points can be plotted as shown below.
The points $P, Q and S$ are on the same line, which makes them collinear.
Therefore, their slopes are equal.
$\begin{aligned} \Rightarrow \frac{2}{t_{1} - t_{2}} = \frac{2}{{t_{1}}^{2} - 1} \\ \Rightarrow {t_{1}}^{2} - 1 = {t_{1}}^{2} + t_{1} t_{2} \\ \Rightarrow t_{1} t_{2} = - 1 \end{aligned}$
It cannot be denied that $t_{1} t_{2} = - 1$ .
Hence option (C) is the answer

Suggest Corrections

Similar questions

t_{1}

t_{2}

y^{2} = 4 a x

t_{1} t_{2} =

If $t_{1} a n d t_{2}$ are end of a focal chord of the parabola $(y^{2} = 4 a x)$ then $t_{1} . t_{2} = - 1$

t_{1}

t_{2}

y^{2} = 4 a x

y^{2} = 4 a x

(a t_{1}^{2}, 2 a t_{1})

(a t_{2}^{2}, 2 a t_{2})

t_{1} . t_{2} = - 1

y^{2} = 4 a x

Join BYJU'S Learning Program