Let A and B be two points on a parabola y2-x with vertex V such that VA is perpendicular to VB and - is the angle between the chord VA and the axis of the parabola- The value of -VA-VB- is-

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Area of a Triangle

Let A and ...

Question

Let $A$ and $B$ be two points on a parabola $y^{2} = x$ with vertex $V$ such that $V A$ is perpendicular to $V B$ and $θ$ is the angle between the chord $V A$ and the axis of the parabola. The value of $\frac{| V A |}{| V B |}$ is:

tan θ

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{tan}^{3} θ

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{cot}^{2} θ

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{cot}^{3} θ

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Solution

The correct option is D ${cot}^{3} θ$
$y^{2} = x$ , vertex is at the $o r i g i n$ and axis is $x - a x i s$ .
Let A has co-ordinates $(a, b)$

Now, according to question
$t a n θ = \frac{b}{a}$

Squaring both sides we get,

$⟹ t a n^{2} θ = \frac{b^{2}}{a^{2}}$

we know that $b^{2} = a$

Hence, we get

$⟹ t a n^{2} θ = \frac{1}{a}$

$⟹ a = c o t^{2} θ$

$⟹ b = c o t θ$

Similarly, for B

Its co-ordinates are { $c o t (90 - θ), c o t^{2} (90 - θ)$ } as $∠ A O B = 90^{\circ}$

$⟹ a = t a n^{2} θ$

$⟹ b = t a n θ$

Now, applying distance formula we get

$\frac{(∣ V A ∣)^{2}}{(∣ V B ∣)^{2}} = \frac{c o t^{4} θ + c o t^{2} θ}{t a n^{4} θ + t a n^{2} θ}$

$\frac{(∣ V A ∣)^{2}}{(∣ V B ∣)^{2}} = \frac{c o t^{2} θ (c o t^{2} θ + 1)}{t a n^{4} θ (c o t^{2} θ + 1)}$

$\frac{(∣ V A ∣)^{2}}{(∣ V B ∣)^{2}} = c o t^{6} θ$

$\frac{(∣ V A ∣)}{(∣ V B ∣)} = c o t^{3} θ$

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Let A and B be two points on a parabola y2=x with vertex V such that VA is perpendicular to VB and θ is the angle between the chord VA and the axis of the parabola. The value of |VA||VB| is:

Let $A$ and $B$ be two points on a parabola $y^{2} = x$ with vertex $V$ such that $V A$ is perpendicular to $V B$ and $θ$ is the angle between the chord $V A$ and the axis of the parabola. The value of $\frac{| V A |}{| V B |}$ is: