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Standard XII

Mathematics

Position of a Line W.R.T Hyperbola

Let PQ be a f...

Question

Let PQ be a focal chord of the parabola $y^{2} = 4 x$ . The tangents to the parabola at P and Q meet at a point on
y = 2x + 1.

If PQ subtends an angle $θ$ at the vertex of $y^{2} = 4 x$ , then $t a n θ$ =

- \frac{2 \sqrt{5}}{3}

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\frac{2 \sqrt{5}}{3}

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\frac{2 \sqrt{7}}{3}

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- \frac{2 \sqrt{7}}{3}

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Solution

The correct option is A $- \frac{2 \sqrt{5}}{3}$
$P = (t^{2}, 2 t), Q = (\frac{1}{t^{2}}, \frac{- 2}{t}) ∴ t - \frac{1}{t} = - 1 \Rightarrow t + \frac{1}{t} = \sqrt{5} \to P Q = a {(t + \frac{1}{t})}^{2} = 5 \to t a n θ = \frac{2 (\frac{1}{t} + t)}{1 - 4} = \frac{2 \sqrt{5}}{3}$

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Similar questions

Let $P Q$ be a focal chord of $y^{2} = 4 a x$ . The tangents to parabola at $P$ and $Q$ meet at a point lying on the line $y = 2 x + a (a > 0)$ . If the chord $P Q$ subtends an angle $θ$ at the vertex of prabola then $tan θ =$

Q. Let

P Q

be a focal chord of the parabola

y^{2} = 4 a x

. The tangents to the parabola at

P

and

Q

meet at a point lying on the line

y = 2 x + a, a > 0

.
If chord

P Q

subtends an angle

θ

at the vertex of

y^{2} = 4 a x

, then tan

θ

is equal to

Q. Let PQ be a focal chord of the parabola

y^{2} = 4 x

. The tangents to the parabola at P and Q meet at a point on
y = 2x + 1.

Length of chord PQ is

Q. Let

P Q

be a focal chord of the parabola

y^{2} = 4 a x

. The tangents to the parabola at

P

and

Q

meet at a point lying on the line

y = 2 x + a, a > 0

.
If chord

P Q

subtends an angle

θ

at the vertex of

y^{2} = 4 a x

, then tan

θ

is equal to

Q. Let

P Q

be a focal chord of the parabola

y^{2} = 4 a x .

The tangents to the parabola at

P

and

Q

meet at a point lying on the line

y = 2 x + a,

a > 0.

If chord

P Q

subtends an angle

θ

at the vertex of

y^{2} = 4 a x,

then

tan θ =

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Let PQ be a focal chord of the parabola y2=4x. The tangents to the parabola at P and Q meet at a point on y = 2x + 1. If PQ subtends an angle θ at the vertex of y2=4x, then tanθ =

The correct option is A −2√53P=(t2,2t),Q=(1t2,−2t)∴t−1t=−1⇒t+1t=√5→PQ=a(t+1t)2=5→tanθ=2(1t+t)1−4=2√53

Let PQ be a focal chord of the parabola $y^{2} = 4 x$ . The tangents to the parabola at P and Q meet at a point on
y = 2x + 1.

If PQ subtends an angle $θ$ at the vertex of $y^{2} = 4 x$ , then $t a n θ$ =

The correct option is A $- \frac{2 \sqrt{5}}{3}$
$P = (t^{2}, 2 t), Q = (\frac{1}{t^{2}}, \frac{- 2}{t}) ∴ t - \frac{1}{t} = - 1 \Rightarrow t + \frac{1}{t} = \sqrt{5} \to P Q = a {(t + \frac{1}{t})}^{2} = 5 \to t a n θ = \frac{2 (\frac{1}{t} + t)}{1 - 4} = \frac{2 \sqrt{5}}{3}$