A curve is represented by the equations x=sec 2 t and y=cot t, where t is a parameter. If the tangent at the point P on the curve where t=π / 4 meets the curve again at the point Q, then |PQ| is equal to

Question

A curve is represented by the equations x=sec 2 t and y=cot t, where t is a parameter. If the tangent at the point P on the curve where t=π / 4 meets the curve again at the point Q, then |PQ| is equal to (A) (5 √3/2) (B) (5 √5/2) (C) (2 √5/3) (D) (3 √5/2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/1ad94abd8d9e6047bc263baaf3f46a53-.png" /> Eliminating t gives y2(x-1)=1 Equation of the tangent at P(2,1) is x+2 y=4, Solving with curve x=5 and y=-1 / 2, we get Q ≡(5,-1 / 2) or PQ=(3 √5/2)

Q. A curve is represented by the equations $x = se c^{2} t$ and $y = co t t$ , where $t$ is a parameter. If the tangent at the point $P$ on the curve where $t = π /4$ meets the curve again at the point $Q$ , then $∣ PQ ∣$ is equal to

$\frac{5 3}{2}$

$\frac{5 5}{2}$

$\frac{2 5}{3}$

$\frac{3 5}{2}$

Eliminating $t$ gives $y^{2} (x - 1) = 1$
Equation of the tangent at $P (2, 1)$ is $x + 2 y = 4$ ,
Solving with curve $x = 5$ and $y = - 1/2,$ we get
$Q \equiv (5, - 1/2)$ or $PQ = \frac{3 5}{2}$

Q. A curve is represented by the equations x=sec2t and y=cott, where t is a parameter. If the tangent at the point P on the curve where t=π/4 meets the curve again at the point Q, then ∣PQ∣ is equal to

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