Tangents to the parabola y2=4ax at P(a t12 , 2 a t1) and Q(a t22 , 2 a t2) meet at T. If triangle PTQ is right-angled at T, then (1/P S)+(1/Q S) is equal to (where, S is the focus of the given parabola)

Question

Tangents to the parabola y2=4ax at P(a t12 , 2 a t1) and Q(a t22 , 2 a t2) meet at T. If triangle PTQ is right-angled at T, then (1/P S)+(1/Q S) is equal to (where, S is the focus of the given parabola) (A) (1/a) (B) (2/a) (C) (1/2 a) (D) (1/4 a)

Tardigrade Admin · Accepted Answer

T lies on the directrix and PQ is the focal chord. ⇒ t1t2=-1 (1/P S)=(1/a (1 + t12)) , (1/Q S)=(1/a (1 + t22))=(t12/a (1 + t12)) So, (1/P S)+(1/Q S)=(1/a)

Q. Tangents to the parabola $y^{2} = 4 a x$ at $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$ meet at $T .$ If $△ PTQ$ is right-angled at $T,$ then $\frac{1}{PS} + \frac{1}{QS}$ is equal to (where, $S$ is the focus of the given parabola)

$\frac{1}{a}$

$\frac{2}{a}$

$\frac{1}{2 a}$

$\frac{1}{4 a}$

$T$ lies on the directrix
and $PQ$ is the focal chord.
$\Rightarrow t_{1} t_{2} = - 1$
$\frac{1}{PS} = \frac{1}{a ( 1 + t _{1}^{2} )}$ , $\frac{1}{QS} = \frac{1}{a ( 1 + t _{2}^{2} )} = \frac{t _{1}^{2}}{a ( 1 + t _{1}^{2} )}$
So, $\frac{1}{PS} + \frac{1}{QS} = \frac{1}{a}$

Q. Tangents to the parabola y2=4ax at P(at12​,2at1​) and Q(at22​,2at2​) meet at T. If △PTQ is right-angled at T, then PS1​+QS1​ is equal to (where, S is the focus of the given parabola)

a1​

a2​

2a1​

4a1​