Let any double ordinate P N P' of the hyperbola (x2/25)-(y2/16)=1 be produced both sides to meet the asymptotes in Q and Q', then P Q ⋅ P' Q is equal to

Question

Let any double ordinate P N P' of the hyperbola (x2/25)-(y2/16)=1 be produced both sides to meet the asymptotes in Q and Q', then P Q ⋅ P' Q is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

N P=(4/5)=√x12-25 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/83357863eb916915233bd40724791040-.png" /> Q is on y=(4/5) x N Q=(4/5) x1 P Q=N Q-N P=(4/5)(x1+√x12-25) P' Q=(4/5)(x1+√x12-25) P Q ⋅ P' Q=16 .

Q. Let any double ordinate $PN P^{'}$ of the hyperbola $\frac{x ^{2}}{25} - \frac{y ^{2}}{16} = 1$ be produced both sides to meet the asymptotes in $Q$ and $Q^{'}$ , then $PQ \cdot P^{'} Q$ is equal to

$NP = \frac{4}{5} = x_{1}^{2} - 25$

$Q$ is on $y = \frac{4}{5} x$
$NQ = \frac{4}{5} x_{1}$
$PQ = NQ - NP = \frac{4}{5} (x_{1} + x_{1}^{2} - 25)$
$P^{'} Q = \frac{4}{5} (x_{1} + x_{1}^{2} - 25)$
$PQ \cdot P^{'} Q = 16.$

Q. Let any double ordinate PNP′ of the hyperbola 25x2​−16y2​=1 be produced both sides to meet the asymptotes in Q and Q′, then PQ⋅P′Q is equal to

NP=54​=x12​−25​ Q is on y=54​x NQ=54​x1​ PQ=NQ−NP=54​(x1​+x12​−25​) P′Q=54​(x1​+x12​−25​) PQ⋅P′Q=16.

Q. Let any double ordinate $PN P^{'}$ of the hyperbola $\frac{x ^{2}}{25} - \frac{y ^{2}}{16} = 1$ be produced both sides to meet the asymptotes in $Q$ and $Q^{'}$ , then $PQ \cdot P^{'} Q$ is equal to

$NP = \frac{4}{5} = x_{1}^{2} - 25$

$Q$ is on $y = \frac{4}{5} x$
$NQ = \frac{4}{5} x_{1}$
$PQ = NQ - NP = \frac{4}{5} (x_{1} + x_{1}^{2} - 25)$
$P^{'} Q = \frac{4}{5} (x_{1} + x_{1}^{2} - 25)$
$PQ \cdot P^{'} Q = 16.$