If $t_{1}$ and $t_{2}$ are the ends of a focal chord of the parabola $y^{2} = 2 x$ then

t_{1}^{2} + t_{2}^{2} = 2

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t_{1} + t_{2} = 1

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t_{1} t_{2} = - 1

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Solution

The correct option is C $t_{1} t_{2} = - 1$
Given parabola $y^{2} = 4 a x$
Since, P, S and Q are collinear.
$m_{P Q} = m_{P S}$
$\frac{2}{t_{1} + t_{2}} = \frac{2 t_{1}}{t_{1}^{2} - 1}$
$\Rightarrow t_{1}^{2} - 1 = t_{1}^{2} + t_{1} t_{2}$
$\Rightarrow t_{1} t_{2} = - 1$

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Similar questions

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

Q. If

t_{1}

and

t_{2}

are the extremities of any focal chord of the parabola

y^{2} = 4 a x

then

t_{1} t_{2} =

If the chord joining the points $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{2}^{2}, 2 a t_{2})$ of the parabola $y^{2} = 4 a x$ passes through the focus of the parabola, then

Q. If the coordinates of the Parabola of the ends of a focal chord of Parabola

y^{2} = 4 a x

are

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

and

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

, then prove that :

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

Q. If the tangents at

t_{1}

and

t_{2}

to a parabola are perpendicular then the value of

| (t_{1} + t_{2})^{2} - (t_{1} - t_{2})^{2} |

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If t1 and t2 are the ends of a focal chord of the parabola y2=2x then

The correct option is C t1t2=−1Given parabola y2=4axSince, P, S and Q are collinear.mPQ=mPS2t1+t2=2t1t21−1⇒t21−1=t21+t1t2⇒t1t2=−1

If $t_{1}$ and $t_{2}$ are the ends of a focal chord of the parabola $y^{2} = 2 x$ then

The correct option is C $t_{1} t_{2} = - 1$
Given parabola $y^{2} = 4 a x$
Since, P, S and Q are collinear.
$m_{P Q} = m_{P S}$
$\frac{2}{t_{1} + t_{2}} = \frac{2 t_{1}}{t_{1}^{2} - 1}$
$\Rightarrow t_{1}^{2} - 1 = t_{1}^{2} + t_{1} t_{2}$
$\Rightarrow t_{1} t_{2} = - 1$