The Line That Is Normal To The Curve X^2 2xy-3y^2=0 At The Point 1 1 Intersects The Curve At What

At what point(s) does the line normal to the curve y=x^2 -3x +5 at the point (3,5) intersect the curve?

y = x^2 - 3x + 5

Slope of tangent to y = y' = 2x - 3
At point (3, 5), x = 3, so y' = 2*3 - 3 = 3

Slope of normal at (3, 5) = m = -1/3 (using the perpendicular slopes formula)
Equation of normal at (3, 5) is y - y1 = m(x - x1), i.e. y = (-1/3)(x - 3) + 5

Equation of normal intersects y = x^2 - 3x + 5 when :
(-1/3)(x - 3) + 5 = x^2 - 3x + 5
i.e. when 3x^2 - 8x - 3 = 0
Factorise: (x - 3)(3x + 1) = 0
x = 3 or -1/3

If x = -1/3, then y = x^2 - 3x + 5 = (-1/3)^2 - 3(-1/3) + 5 = 55/9
Thus, the other point of intersection is (-1/3, 55/9).

Normal line intersecting a curve?

First, we would desire to locate the conventional vector. this may well be a line commonly used to the tangent at (a million,0). to locate the slope of this tangent, take the by-product at (a million,0) y = x^2 - x y' = 2x - a million y'(a million) = 2*a million - a million = a million So in y = mx + b, m = a million. Now locate b: 0 = a million*a million + b b = a million. So the tangent line is y = x - a million A line commonly used to this, intersecting at (a million,0) may well be: y = -x + a million Now, the standards the place this line intersects the parabola are the standards while -x + a million = x^2 - x. x^2 -a million = 0 x = a million, x = -a million We already understand approximately x = a million, when you consider that we've been given (a million,0). Plug in x = -a million to get: y = (-a million)^2 - (-a million) = 2. So it additionally intersects at (-a million,2)

The line that is normal to the curve x^2 + xy - 2y^2 = 0 at (4,4) intersects the curve at what other point?

f(x) = x^2 + xy - 2y^2
f'(x) = 2x + (y + xy') - 4yy'

At (4,4): 8 + (4 + 4y') - 16y' = 0 --> 12 = 12y' ---> y' = 1
==> normal slope: -1
Through (4,4)
y -4 = -1( x -4)
y = -x + 8

Intersects points between normal line and curve:
x^2 + x(-x+8) - 2(-x + 8)^2 = 0
x^2 - x^2 + 8x -2 (x^2 - 16x + 64) = 0
-2x^2 + 40x - 128 = 0
x^2 - 20x + 64 = 0
delta = 100 - 64 = 36
x = 10 +/- 6 ==> x = 4 and x= 16
y = -x + 8 = 16 + 8 = 24
==> Another point (16, 24)

The line that is normal to the curve x^2+xy-2y^2=0 at (4,4) intersects the curve at what other point.?

the slope of the curve is equal the the first derivative. The normal of the curve is perpendicular to the tangent line, and also goes though the tangent point.

x^2+ xy − 2y^2 = 0
∴ d/dx(x^2+ xy − 2y^2) = d/dx(0)
term by term
∴ d/dx(x^2)+ d/dx(xy) − d/dx(2y^2) = d/dx(0)
using the Product rule d/dx(uv) = u * dv/dx + v * du/dx
∴ d/dx(x^2)+ (x * dy/dx + y * d/dx(x)) − d/dx(2y^2) = d/dx(0)
using the Chain rule d/dx(u^n) = nu^(n−1) * du/dx
∴ d/dx(x^2)+ (x * dy/dx + y * d/dx(x)) − 4y * dy/dx = d/dx(0)
using the Power rule d/dx(x^n) = nx^(n−1)
∴ 2x + (x * dy/dx + y * 1) − 4y * dy/dx = 0
∴ 2x + x(dy/dx) + y − 4y(dy/dx) = 0
∴ x(dy/dx) − 4y(dy/dx) = -2x − y
∴ (dy/dx)(x − 4y) = -(2x + y)
∴ dy/dx = -(2x + y)/(x − 4y)

the slope of the tangent at (4,4) = -(2*4 + 4)/(4 − 4*4) = -12/(-12) = 1
the slope of the normal at (4,4) = -1/(1) = -1

using the point slope formula y − y₁ = m(x − x₁)
∴ y − 4 = -1(x − 4)
∴ y − 4 = -x + 4
∴ y = -x + 8 <== the normal of the curve at (4,4)

now it also passes though
x^2+ xy − 2y^2 = 0
∴ x^2 + x(8 − x) − 2(8 − x)^2 = 0
∴ x^2 + 8x − x^2 − 2(64 − 16x + x^2) = 0
∴ x^2 + 8x − x^2 − 128 + 32x − 2x^2 = 0
∴ 2x^2 − 40x + 128 = 0
∴ 2x^2 − 8x − 32x + 128 = 0
∴ 2x(x − 4) − 32(x − 4) = 0
∴ (x − 4)(2x − 32) = 0
∴ 2(x − 4)(x − 16) = 0
so the normal passes though x = 4 and x = 16, as we already know about (4,4)

y = -16 + 8 = -8

the Normal also pass though (16,-8)

The line that is normal to the curve x^2 + 2xy - 3(y)^2= 0 ...?

x² + 2xy - 3y² = 0 Take the by-product and evalutate it on the ingredient (a million, a million). 2x + 2y + 2x(dy/dx) - 6y(dy/dx) = 0 x + y + x(dy/dx) - 3y(dy/dx) = 0 (x - 3y)(dy/dx) = -x - y dy/dx = (-x - y)/(x - 3y) = (x + y)/(3y - x) dy/dx = (a million + a million)/(3*a million - a million) = 2/2 = a million The slope of the equation of the line well-known to the curve at (a million,a million) is the damaging reciprocal or m = -a million. The equation of the traditional line is: y - a million = -(x - a million) = -x + a million y = -x + 2 _______ Plug the value for y into the equation of the curve and sparkling up for x. x² + 2xy - 3y² = 0 x² + 2x(-x + 2) - 3(-x + 2)² = 0 x² - 2x² + 4x - 3(x² - 4x + 4) = 0 x² - 2x² + 4x - 3x² + 12x - 12 = 0 -4x² + 16x - 12 = 0 x² - 4x + 3 = 0 (x - a million)(x - 3) = 0 x = a million, 3 ______ Plug the value of x into the curve and sparkling up for y. Plug in x = a million. x² + 2xy - 3y² = 0 a million + 2y - 3y² = 0 3y² - 2y - a million = 0 (3y + a million)(y - a million) = 0 y = -a million/3, a million in straight forward terms a sort of values is on the traditional line. the different answer is extraneous. the only answer is: y = a million So one ingredient on the intersection of the curve and the traditional line is (a million, a million). _______ Plug the value of x into the curve and sparkling up for y. Plug in x = 3. x² + 2xy - 3y² = 0 9 + 6y - 3y² = 0 -3 - 2y + y² = 0 y² - 2y - 3 = 0 (y + a million)(y - 3) = 0 y = -a million, 3 in straight forward terms a sort of values is on the traditional line. the different answer is extraneous. the only answer is: y = -a million So a 2nd ingredient on the intersection of the curve and the traditional line is (3, -a million). _______ the two factors of intersection are: (a million, a million) and (3, -a million)

The line that is normal to the curve x^2 + xy - 2y^2 = 0 at (2,2) intersects the curve at what other point?

x^2 + xy - 2y^2 = 0
2x + y + x(dy/dx) - 4y(dy/dx) = 0
(dy/dx)(x - 4y) = -2 - y
dy/dx = (2 + y) / (4y - x)
At (2,2), dy/dx = (2 + 2) / (8 - 2) = 4/6 = 2/3.
So our normal line will have a slope of -3/2 and pass through (2,2).
y - 2 = (-3/2)(x - 2)
y - 2 = -3x/2 + 3
y = -3x/2 + 5
This intersects our curve when:
x^2 + x(-3x/2 + 5) - 2(-3x/2 + 5)^2 = 0
x^2 - 3(x^2)/2 + 5x - 2(9(x^2)/4 - 15x + 25) = 0
x^2 - 3(x^2)/2 + 5x - 9(x^2)/2 + 30x - 50 = 0
2x^2 - 3x^2 + 10x - 9x^2 + 60x - 100 = 0
-10x^2 + 70x - 100 = 0
x^2 - 7x + 10 = 0
x^2 - 2x - 5x + 10 = 0
x(x - 2) - 5(x - 2) = 0
(x - 2)(x - 5) = 0
=> x - 2 = 0, x = 2 (this is at the point we already have, (2,2))
=> x - 5 = 0, x = 5
When x = 5, we have y = -15/2 + 5 = -5/2.
So the normal line also intersects the curve at (5,-5/2).

The line that is normal to the curve x^2+2xy-3y^2=0 at the point (1,1) intersects the curve at what other point?

x^2 + 2x*y - 3y^2 = 0
2x + 2xdy/dx + y*2 – 6ydy/dx = 0
(6y – 2x)dy/dx = 2x + 2y
dy/dx = (x + y)/(3y – x)
At the point (1, 1) the slope is (1 + 1)/(3 – 1) = 1
The slope of the normal is -1 so line is y = 2 – x and substituting
x^2 + 2x(2 – x) – 3(2 – x)^2 = -4x^2+ 16x – 12 = 0
-4(x – 1)(x – 3) = 0
The other point has x = 3
Substituting that yields 9 + 6y – 3y^2 = 0

Wondering why it seemed that y could take 2 values, 3 or -1.
Plotted original equation. Found that it represents two intersecting lines.
Specifying point (1, 1) selected one of the lines, so the slope was appropriate
The graphs inform us that the correct choice for the second point is (3, - 1)

Note: The point (3, 3) is further along the line selected by (1, 1).
Can anyone suggest a geometric interpretation for why this other option occurs

Regards – Ian H

The line that is normal to the curve x^2 + 2xy - 3y^2 = 0 at (1,1) intersects the curve at what other point?

x² + 2xy - 3y² = 0

Take the derivative and evalutate it at the point (1, 1).

2x + 2y + 2x(dy/dx) - 6y(dy/dx) = 0
x + y + x(dy/dx) - 3y(dy/dx) = 0
(x - 3y)(dy/dx) = -x - y
dy/dx = (-x - y)/(x - 3y) = (x + y)/(3y - x)
dy/dx = (1 + 1)/(3*1 - 1) = 2/2 = 1

The slope of the equation of the line normal to the curve at (1,1) is the negative reciprocal or m = -1. The equation of the normal line is:

y - 1 = -(x - 1) = -x + 1
y = -x + 2
_______

Plug the value for y into the equation of the curve and solve for x.

x² + 2xy - 3y² = 0
x² + 2x(-x + 2) - 3(-x + 2)² = 0
x² - 2x² + 4x - 3(x² - 4x + 4) = 0
x² - 2x² + 4x - 3x² + 12x - 12 = 0
-4x² + 16x - 12 = 0
x² - 4x + 3 = 0
(x - 1)(x - 3) = 0
x = 1, 3
______

Plug the value of x into the curve and solve for y. Plug in x = 1.

x² + 2xy - 3y² = 0
1 + 2y - 3y² = 0
3y² - 2y - 1 = 0
(3y + 1)(y - 1) = 0
y = -1/3, 1

Only one of these values is on the normal line. The other solution is extraneous. The only solution is:

y = 1

So one point at the intersection of the curve and the normal line is (1, 1).
_______

Plug the value of x into the curve and solve for y. Plug in x = 3.

x² + 2xy - 3y² = 0
9 + 6y - 3y² = 0
-3 - 2y + y² = 0
y² - 2y - 3 = 0
(y + 1)(y - 3) = 0
y = -1, 3

Only one of these values is on the normal line. The other solution is extraneous. The only solution is:

y = -1

So a second point at the intersection of the curve and the normal line is (3, -1).
_______

The two points of intersection are:
(1, 1) and (3, -1)