Question

two tangents are drawn from the point -2,1 to the parabola y2 =4x if theta is angle between tangent then tan theta
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Answer 1

Step-by-step explanation:

Answer:

tan θ = 3

Step-by-step explanation:

Given:

Two tangents are drawn from the point (-2,1)

Equation of the parabola = y² = 4x

θ is the angle between the tangents

To Find:

tan θ

Solution:

Here equation of the parabola is of the form y² = 4ax = y² = ax

Comparing we get the value of a as 1.

Now equation of tangent to a parabola is given by,

Equation of tangent = y = mx + a/m

where m is the slope of the tangent

Here a = 1, hence,

Equation of tangent = y = mx + 1/m

Given that the tangents are drawn from the point (-2, 1) ie, this point lies on the tangent.

Therefore,

1 = -2m + 1/m

Cross multiplying,

1m = -2m² + 1

2m² + m - 1 = 0

Factorising by splitting the middle term,

2m² -m + 2m - 1 = 0

m (2m - 1) + 1 (2m - 1) = 0

(2m - 1) (m + 1) = 0

Either

m + 1 = 0

m = -1

Or

2m - 1 = 0

m = 1/2

Hence the slopes of the tangent are -1 and 1/2

Let us take them as m₁ and m₂

Now tan θ between two lines is given by,

$\tt tan\: \theta=\bigg | \dfrac{m_2-m_1}{1+m_1m_2}\bigg|$

Substitute the data,

$\tt tan\: \theta=\dfrac{\dfrac{1}{2} +1}{1+\dfrac{1}{2}\times -1}$

$\tt tan\: \theta=\dfrac{3}{2}\div \dfrac{1}{2}$

$\tt tan\: \theta=3$

Hence the value of tan θ is 3.

Answer 2

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Given:-

Two tangents are drawn from the point (-2,1)

Equation of the parabola = y² = 4x

θ is the angle between the tangents

To Find:-

tan θ

Solution:-

Here equation of the parabola is of the form y² = 4ax = y² = ax

Comparing we get the value of a as 1.

Now equation of tangent to a parabola is given by,

Equation of tangent = y = mx + a/m

where m is the slope of the tangent

Here a = 1, hence,

Equation of tangent = y = mx + 1/m

Given that the tangents are drawn from the point (-2, 1) ie, this point lies on the tangent.

Therefore,

1 = -2m + 1/m

Cross multiplying,

1m = -2m² + 1

2m² + m - 1 = 0

Factorising by splitting the middle term,

2m² -m + 2m - 1 = 0

m (2m - 1) + 1 (2m - 1) = 0

(2m - 1) (m + 1) = 0

Either

m + 1 = 0

m = -1

Or

2m - 1 = 0

m = 1/2

Hence the slopes of the tangent are -1 and 1/2

Let us take them as m₁ and m₂

Now tan θ between two lines is given by,

$\tt tan\: \theta=\bigg | \dfrac{m_2-m_1}{1+m_1m_2}\bigg|tanθ=∣∣∣∣∣1+m1m2m2−m1∣∣∣∣∣</p><p>Substitute the data,</p><p>\tt tan\: \theta=\dfrac{\dfrac{1}{2} +1}{1+\dfrac{1}{2}\times -1}tanθ=1+21×−121+1</p><p>\tt tan\: \theta=\dfrac{3}{2}\div \dfrac{1}{2}tanθ=23÷21</p><p>\tt tan\: \theta=3tanθ=3$

$\bold \red{Hence \: the \: value \: of \: tan \: θ \: is \: 3.}$