Mathematics Test - 39

To find the coordinates of the foot of the normal to the parabola y² = 5x that makes a 45° angle with the x-axis, follow these steps:

Step 1: Understand the Parabola and Its Normal

The given parabola is:

y² = 5x

This is a standard parabola of the form y² = 4ax, where 4a = 5, so a = 5/4.

Differentiate both sides to find the slope of the tangent:

2y (dy/dx) = 5

So, dy/dx = 5 / (2y)

The slope of the normal (which is perpendicular to the tangent) is the negative reciprocal of that:

Slope of normal = -2y / 5

Step 2: Use the Given Angle of the Normal

We're told the normal makes a 45° angle with the x-axis, so its slope is:

Slope = tan(45°) = 1

Now, equate this to the expression for the slope of the normal:

-2y / 5 = 1
Solving for y gives: y = -5/2

Step 3: Find the Corresponding x-Coordinate

Substitute y = -5/2 into the parabola equation to find x:

(-5/2)² = 5x

25/4 = 5x

Solving for x: x = 5/4

Step 4: Verify the Direction of the Normal

The slope of the normal is 1, which corresponds to a 45° angle with the x-axis.

The point (5/4, -5/2) satisfies this condition.

Step 5: Check the Options

So, the coordinates of the foot of the normal are (5/4, -5/2), which matches Option B.

Given, y² = 3x ……….(1) and y = 2x + 4 ……….(2)

Differentiating both sides of (1) with respect to y we get,

2y = 3(dx/dy)

Or dx/dy = 2y/3

Let P (x₁, y₁) be any point on the parabola (1). Then the slope of the normal to the parabola (1) at point P is

-[dx/dy]_P = -2y₁/3

If the normal at the point P to the parabola (1) be perpendicular to the line (2) then we must have,

-2y₁/3*2 = -1

Since the slope of the line (2) is 2

Or y₁ = 3/4

Since the point P(x₁, y₁) lies on (1) hence,

y₁² = 3x₁

As, y₁ = 3/4, so, x₁ = 3/16

Therefore, the required equation of the normal is

y – y₁ = -(2y₁)/3*(x – x₁)

Putting the value of x₁ and y₁ in the above equation we get,

16x + 32y = 27

And the coordinates of the foot of the normal are (x₁, y₁) = (3/16, 3/4)

Given, x² + 3y² = 12 Or x²/12 + y²/4 = 1

Differentiating both sides of (1) with respect to y we get,

2x*(dx/dy) + 3*2y = 0

Or dx/dy = -3y/x

Suppose the normal to the ellipse (1) at the point P(√12cosθ, 2sinθ) makes an angle 60° with the major axis. Then, the slope of the normal at P is tan60°

Or -[dx/dy]_P = tan60°

Or -(-(3*2sinθ)/√12cosθ) = √3

Or √3tanθ = √3

Or tanθ = 1

Now the centre of the ellipse (1) is C(0, 0)

Therefore, the slope of the line CP is,

(2sinθ – 0)/(√12cosθ – 0) = (1/√3)tanθ = 1/√3 [as, tanθ = 1]

Therefore, the line CP is inclined at 30° to the major axis.

x² – y² = 2a² ……….(1) and x² + y² = 4a² ……….(2)

Adding (1) and (2) we get, 2x² = 6a²

Again, (2) – (1) gives,

2y² = 2a²

Therefore, 2x² * 2y² = 6a² * 2a²

4x²y² = 12a²

Or x²y² = 3a⁴

Or 2xy = ±2√3

Differentiating both side of (1) and (2) with respect to x we get,

2x – 2y(dy/dx) = 0

Or dy/dx = x/y

And 2x + 2y(dy/dx) = 0

Or dy/dx = -x/y

Let (x, y) be the point of intersection of the curves(1) and (2) and m₁ and m₂ be the slopes of the tangents to the curves (1) and (2) respectively at the point (x, y); then,

m₁ = x/y and m₂ = -x/y

Now the angle between the curves (1) and (2) means the angle between the tangents to the curve at their point of intersection.

Therefore, if θ is the required angle between the curves (1) and (2), then

tanθ = |(m₁ – m₂)/(1 + m₁m₂)|

Putting the value of m₁, m₂ in the above equation we get,

tanθ = |2xy/(y² – x²)|

As, 2xy = ±2√3a² and x² – y² = 2a²

tanθ = |±2√3a²/-2a²|

Or tanθ = √3

Thus, θ = π/3.

The above is a condition for a symmetric relation.

For example, a relation R on set A = {1,2,3,4} is given by R={(a,b):a+b=3, a>0, b>0}

1+2 = 3, 1>0 and 2>0 which implies (1,2) ∈ R.

Similarly, 2+1 = 3, 2>0 and 1>0 which implies (2,1)∈R. Therefore both (1, 2) and (2, 1) are converse of each other and is a part of the relation. Hence, they are symmetric.

Step 1: Understand the Range of sin⁻¹

The inverse sine function, sin⁻¹(x), gives a value in the range [-π/2, π/2], which is approximately [-1.57, 1.57] radians (or [-90°, 90°]).

Step 2: Reduce the Angle to the Principal Range

We are given the angle 6 radians. Since 6 is not in the principal range of sin⁻¹, we need to find an equivalent angle θ in the range [-π/2, π/2] such that sin(θ) = sin(6).

Step 2.1: Use Periodicity of Sine

The sine function is periodic with a period of 2π. So,

sin(6) = sin(6 - 2π)

Now calculate:

2π ≈ 6.2832

6 - 2π ≈ 6 - 6.2832 = -0.2832 radians

This angle, -0.2832, lies within the required range [-π/2, π/2].

Step 2.2: Verify It's the Only Correct Value

Since -0.2832 is in the correct range and gives the same sine value as 6, we conclude:

sin⁻¹(sin 6) = 6 - 2π

Step 3: Final Answer

• In exact form: sin⁻¹(sin 6) = 6 - 2π

• In decimal form: ≈ -0.2832 radians

Poisson distribution shows the number of times an event is likely to occur within a specified time. The Poisson distribution probability formula is P(x; μ) = (e^-μ) (μ^x) / x!

Formula for binomial distribution is P [X = x] = ⁿC_x p^x q^n-x

Where n is the number of trials, x is the number of successes and (n-x) is failures.

Bernoulli trials is termed after swiss mathematician Jacob Bernoulli. Bernoulli trials is also called a Dichotomous experiment and is repeated n times. If in each trial the probability of success is constant, then such trials are called Bernoulli trials.

Bernoulli trial has only two possible outcomes and is mutually exclusive. Those two outcomes are ‘success’ and ‘failure’. So, it is also called as a ‘yes’ or ‘no’ question.