#### The diagram above shows the curve with equation y^2 = 4(x - 2) and the line with equation 2x – 3y = 12.

The curve crosses the x-axis at the point A, and the line intersects the curve at the points P and Q.

(a) Write down the coordinates of A.

Anonymous

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The curve crosses the x-axis at the point A, and the line intersects the curve at the points P and Q.

(a) Write down the coordinates of A.

Sangeetha Pulapaka

1

a) The equation of the curve/parabola given is y^{2} = 4(x-2) .

This is intersecting at a point A on the x-axis.

We know that on y = 0, on the x-axis.

So, plugging in y = 0, in the above equation we get 0 = 4(x - 2)

4x - 8 = 0

\Rightarrow x = 2

So point A has the coordinates (2,0).

b)** Find using algebra the points P and Q.**

The equation of the line given is 2x - 3y = 12

So 2x = 12 + 3y

x = 6 + \frac{3}{2}y

Plugging this back into the equation of the parabola, we get

y^2\ =\ 4\left(6\ +\ \frac{3}{2}y\ -\ 2\right)

y^2=\ 4\left(4\ +\ \frac{3}{2}\ y\right)

y^{2} = 16 + 6y

y^{2} -6y - 16 = 0

Factoring this we get

y^{2} - 8y + 2y - 16 = 0

y(y-8) + 2(y - 8)=0

y = -2, y = 8

Plugging y = 8, in 2x - 3y = 12, solve for x, to get,

2x = 12 + 3(8)

\Rightarrow x = 18

Plugging in y = -2, in 2x - 3y = 12, solve for x, to get,

2x = 12 + 3(-2)

2x = 6

\Rightarrow x = 3

So, the coordinates of P and Q are (18 ,8) and (3, -2)

c) Show that \angle PAQ is a right angle.

We have points A(2,0) P(18,8), Q(3,-2)

Recall that when the slopes are perpendicular (or when are at an angle of 90^\circ), the product of the slopes is equal to -1.

Slope of AP can be found by using the slope formula \frac{y_{2}-y_{1}}{x_{2}-x_{1}}

Plugging in the coordinates of AP we get the slope as \frac{8-0}{18-2}= \frac{8}{16} = \frac{1}{2}

Slope of AQ = \frac{-2 -0}{3-2} = -2

The product of the slopes is \frac{1}{2} \cdot -2 = -1

So \angle PAQ is a right angle.

Krishna

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Step 1: Note down the given equations and observe the diagram (point out the intersection points)

Step 2: Put x or y equal to zero to find the intersection point of curve or line with the axis.

NOTE: If the line or curve intersecting the y-axis then put x = 0 in the equation

If the line or curve intersecting the x-axis then put y = 0 in the equation.

EXAMPLE: The curve y^2 = 4(x - 2) intersecting the x-axis

so, y = 0.

0 = 4(x - 2)

Step 3: Solve the equation

NOTE: Apply the inverse operations

0 = 4(x - 2)

(x - 2) = 0

x = 2

y-coordinate "0"

x-coordinate "2"

Intersection points (0, 2)

Anonymous

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Can you answer the second part of that question?

Anonymous

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b-) Find, using algebra, the coordinates of P and Q.

Sangeetha Pulapaka

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Please, look at the completed explanation now!

Sangeetha Pulapaka

0

In case you are wondering, we usually do respond to a comment immediately. It does not take this long :)