If chord PQ subtends an angle - at the vertex of y2-4ax- then tan-

The correct option is D 2/3√(5) Angle made by chord PQ at vertex (0,0) is given by tanθ= (2t+2t/1 4) = #160;2(1t+t) 3= 2/3√(5) ...

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Byju's Answer

Standard XII

Mathematics

Parametric Form of Tangent: Hyperbola

If chord PQ...

Question

If chord $P Q$ subtends an angle $θ$ at the vertex of $y^{2} = 4 a x$ , then $tan θ =$

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

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

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

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

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Solution

The correct option is D $\frac{- 2}{3} \sqrt{5}$
Angle made by chord $P Q$ at vertex $(0, 0)$ is given by
$tan θ = ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{2}{t} + 2 t}{1 - 4} ⎞ ⎟ ⎟ ⎟ ⎠$ $= \frac{2 (\frac{1}{t} + t)}{- 3} = \frac{- 2}{3} \sqrt{5}$

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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.

If PQ subtends an angle

θ

at the vertex of

y^{2} = 4 x

, then

t a n θ

Q. The length of normal chord of parabola

y^{2} = 4 x

, which subtends an angle of

90^{\circ}

at the vertex 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

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subtends an angle

θ

at the vertex of

y^{2} = 4 a x

, then tan

θ

is equal to

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If chord PQ subtends an angle θ at the vertex of y2=4ax, then tanθ=

The correct option is D −23√5Angle made by chord PQ at vertex (0,0) is given bytanθ=⎛⎜ ⎜ ⎜⎝2t+2t1−4⎞⎟ ⎟ ⎟⎠ =2(1t+t)−3=−23√5

If chord $P Q$ subtends an angle $θ$ at the vertex of $y^{2} = 4 a x$ , then $tan θ =$

The correct option is D $\frac{- 2}{3} \sqrt{5}$
Angle made by chord $P Q$ at vertex $(0, 0)$ is given by
$tan θ = ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{2}{t} + 2 t}{1 - 4} ⎞ ⎟ ⎟ ⎟ ⎠$ $= \frac{2 (\frac{1}{t} + t)}{- 3} = \frac{- 2}{3} \sqrt{5}$