# Find coordinates of the point of intersection of the two normals to the curve y=x^2 +3x+5 which make an angle of 45 degree with the x-axisThis is a differentiation question. I don't know how to...

Find coordinates of the point of intersection of the two normals to the curve y=x^2 +3x+5 which make an angle of 45 degree with the x-axis

This is a differentiation question. I don't know how to find the point P1 and P2 -> i can't not generate the normals equation -> can't solve them simultaneously -> stuck...help

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Since the problem provides the intersection that the two tangents to parabola given by the quadratic equation `y=x^2 +3x+5` make an angle of `45^o` with x axis, hence, the slopes of the tangent lines to parabola are equal to tan `45^o` and tan `135^o` such that:

`m_1 = tan 45^o => m = 1`

`m_2 = tan 135^o => m = -1`

You need to remember that the derivative of the function at a point is equal to the slope of the tangent line at the point, such that:

`f'(x_1) = m_1 => f'(x_1) = 2x_1 + 3 => 2x_1 + 3 = 1`

`2x_1 = 1 - 3 => 2x_1 = -2 => x_1 = -1 => y_1 = 1-3+5 = 3`

`f'(x_2) = m_2 => 2x_2 + 3 = -1 => 2x_2 = -4 => x_2 = -2 => y_2 = 3`

You need to remember that the product of slopes of two perpendicular lines is -1 such that:

`m_1*m'_1 = -1 => 1*m'_1 = -1 => m'_1 = -1`

`m_2*m'_2 = -1 => -1*m'_2 = -1 => m'_2 = 1`

You need to write the equation of first normal to to parabola at the point `(-1,3)` such that:

`y - 3 = m'_1(x + 1) => y = 3 - x - 1 => y = -x + 2`

You need to write the equation of second normal to to parabola at the point `(-2,3)` such that:

`y - 3 = m'_2(x +2) => y = x + 5`

You need to find the intersection point between the two normals, hence, you need to solve the system of simultaneous equations such that:

`{(y = -x + 2),(y = x + 5):} => -x + 2 = x + 5`

You need to isolate the terms that contain x to the left side, such that:

`-x - x = 5 - 2 => -2x = 3 => x = -3/2`

You need to find y coordinate such that:

`y = -3/2 + 5 => y = 7/2`

**Hence, evaluating the coordinates of the point of intersection of the normals to the given parabola, yields `(-3/2,7/2).` **

tan 45 degree= 1 or -1

When y' = 1 (y' = 2x + 3 =1)

x = -1

-> y =(-1)^2+3(-1)+5

->y= 3

**Point 1 (-1,3)**

When y' = -1 (y' = 2x + 3 =-1)

x = -2

-> y =(-2)^2+3(-2)+5

->y= 3

**Point 2 (-2,3)**

**The equation of tangent 1 at point 1 is:**

y - 3 = -1(x + 1)

y - 3 = -x -1

y= -x + 2 * (1)*

**The equation of tangent 2 at point 2 is:**

y - 3 = 1(x + 2)

y - 3 = x+2

y = x + 5 (2)

Solving simultaneosly *(1)(2)* we have:

X = (7/2) Y=(3/2)

(7/2, 3/2)

However, the final answer is (-3/2,7/2). Anyone figured out why???