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# how do you get an equation of a tangent line to a parabola given the point of tangency

given a point of tangency on a parabola, how do you come up with the equation fot the tangent lin?

## Research, Knowledge and Information :

### How to Find the Tangent Lines of a Parabola that ... - dummies

Ever want to determine the location of a line through a given point that’s tangent to a given ... tangent line at that point. So, you ... equation of the parabola ...

### How to get the equation of a tangent to a parabola from a ...

You are given the point ... How can I get the equation of a tangent to a parabola from a point that ... find the equation for the line and point of tangency. if ...

### Equation of a Line, Tangent Lines - Brown ... - Brown University

Equation of a Line, Tangent Lines. ... It is very easy to find the equation for the tangent line to a curve at a ... to find the tangent line at a point x 0, you need ...

### Tangent of Parabola Algebraically - Mathwarehouse.com

Tangent of Parabola Algebraically ... Therefore the equation of a tangent line through any point on the parabola y =x 2 has a slope of 2x

### Find the equation of the parabola given the tangent to a ...

Find the equation of the parabola given the tangent to a point and another point. ... Find an equation of the tangent line to the parabola passing through \$ ...

### ︎² How to Find the Equation of a Tangent Line with ...

Jul 27, 2014 · MIT grad shows how to find the equation of a tangent line ... the point. You get this by plugging the given x ... the equation of a tangent to ...

### Tangent at a Given Point on the Standard Parabola - tpub.com

TANGENT AT A GIVEN POINT ON THE STANDARD PARABOLA. The standard parabola is represented by the equation. Let P 1 with coordinates (x 1,y 1) be a point on the curve.

### How to Find the Equation of a Tangent Line: 8 Steps

Apr 03, 2016 · How to Find the Equation of a Tangent Line. ... write the tangent line equation in point ... given the slope of the tangent line, you'll need to ...

### equation of tangent to a parabola - YouTube

Feb 24, 2009 · Calculation of the equation of a tangent to a parabola at a given point, using the derivative of the function and simple gradient formula for two points

## Suggested Questions And Answer :

### how do you get an equation of a tangent line to a parabola given the point of tangency

You need to find the derivative dy/dx of the curve and substitute x and y, if necessary, to find the slope at a particular given point. The slope is the tangent at the point. The standard linear equation y=ax+b can then be constructed, because the tangent will be a. To find b, the y intercept, substitute the coords of the given point into x and y and you'll be able to find b. That gives you the line. For example, let y=px^2+qx+r be the equation of a parabola, where p, q and r are constants, then dy/dx=2px+q. Let's say you're given the point on the curve where x=A, then dy/dx=2pA+q. This is a number because p, q and A are all known values (you'll be told what they are). This is your slope for the tangent, so a=2pA+q, and y=(2pA+q)x+b is the equation of the tangent line. To find b we first find out what y is on the curve when x=A. We know y=pA^2+qA+r. To make it easier to read, call this y coord B. So we substitute (A,B) in our linear equation to find b=B-A(2pA+q). It's much easier when you have actual numbers rather than symbols. In the same way you can find the equation of the perpendicular because its slope is related to the slope of the tangent. It's -1/(2pA+q), the negative reciprocal of the tangent. I hope this helps your understanding.

### Find an equation of the tangent line

The intersection of the line and curve can be found by substituting y=-(4+2x) into the first equation: 16+16x+4x^2+16+8x-8x-20=0; 4x^2+16x+12=0; x^2+4x+3=0; (x+1)(x+3)=0, so x=-1 and -3. The y values for these values are y=-(4-2)=-2 and -(4-6)=2. So the intersection points are; (-1,-2) and (-3,2). Differentiating: 2ydy/dx-4dy/dx-8=0. At (-1,-2) the gradient is given by dy/dx in: -4dy/dx-4dy/dx-8=0; -8dy/dx=8, so dy/dx=-1; at (-3,2), however, 4dy/dx-4dy/dx-8=0, and dy/dx cancels out, leaving the inequality -8=0, so there is apparently only one useful intersection, and the gradient is -1. The tangent line is of the form y=mx+c, where m=dy/dx, the slope, is -1. To find c, insert the point (-1,-2): -2=1+c, making c=-3 and y=-x-3 is the tangent line.  The tangent at (-3,2) is infinity, which explains why we can't get a gradient. The tangent line is, in fact, the line x=-3, and the general equation for this line cannot be covered by the standard equation y=mx+c, because x=-3 for all values of y. So there are two tangent lines: y=-3-x at (-1,-2) and x=-3 at (-3,2). The curve is a sideways parabola with vertex at (-3,2), where the gradient is infinite (the angle is 90 degrees).

### line d y=2x+4 ,e y=2x+c find c and the lines are parallel and have 6 units between them

line d y=2x+4 ,e y=2x+c find c and the lines are parallel and have 6 units between them There are two lines that satisfy the given parameters, one above and to the left of line d, and one below and to the right of line d. To find a point 6 units away from line d, we need a line that is perpendicular to d. We get the equation of that line by using the negative reciprocal of the m component, 2: that gives us -1/2. The equation of this line is y = -1/2 x + 4, using the same y intercept. A segment of that line that is 6 units represents the hypotenuse of a right triangle. If we are looking at the segment that extends to the left of the given y intercept, we consider the distance to be negative (-6). If we are looking at the segment that extends to the right of the given y intercept, we consider the distance to be positive (+6). To find the x and y co-ordinates, we use the sine and the cosine of the angle (we'll call it a) formed by this new line and the x axis. The slope m of a line is the y distance divided by the x distance. That is also the definition of the tangent of the angle. Using the inverse tangent, we determine that the angle is tan^-1 (-1/2) =  -26.565 degrees. We need to keep all the signs straight in order to get the correct values. The sine of -26.565 degrees is -0.4472. The cosine of -26.565 degrees is 0.8944. y1 / -6 = sin a y1 = -6 sin a = -6 * -0.4472 = 2.6832 This is measured from a horizontal line through the y intercept, because we are constructing a right-triangle with one corner at the y intercept, so the point's y co-ordinate is actually y = 2.6832 + 4 = 6.6832 x / -6 = cos a x = -6 * 0.8944 = -5.3664 Making a quick check using x^2 + y^2 = r^2: (-5.3664)^2 + 2.6832^2 = 28.79825 + 7.19956 = 35.99781   -6^2 = 36 Close enough considering the rounding in the calculations. We have co-ordinate (-5.3664, 6.6832) lying 6 units from line d. Substituting those values into y = 2x + c we have 6.6832 = 2 * (-5.3664) + c 6.6832 = -10.7328 + c 6.6832 + 10.7328 = c = 17.416 The equation of a line 6 units away from d and parallel to it, located above it, is y = 2x + 17.416    <<<<<<<< Working in the other direction, on the line below and to the right of line d, y2 / 6 = sin a y2 = 6 * sin a = 6 * -0.4472 = -2.6832 (that is, 2.6832 below d's y intercept) y = 4 - 2.6832 = 1.3168 x / 6 = cos a x = 6 * cos a = 6 * 0.8944 = 5.3664 The co-ordinates for this point, (5.3664, 1.3168), when substituted into the equation y = 2x + f (f is a new y intercept) gives 1.3168 = 2 * 5.3664 + f 1.3168 - 10.7328 = f = -9.416 The equation of a line 6 units away from d and parallel to it, located below it, is y = 2x - 9.416    <<<<<<<<

### what is each part of the equation y=mx+b mean and who do you find them using math vocabulary

The normal meaning for this standard linear equation is that x and y are coordinates in a rectangular arrangement of axes. The y axis is North-South while the x axis is East-West. Where they cross is called the origin with coordinates (0,0), that is, x and y are both zero. The equation y=mx+b defines a straight line. It slopes at a value given by m, the slope or gradient, and m is a number which can be a whole number, a fraction, or whatever, as long as it is constant so that the line remains straight. The slope, m, is also known as the tangent, and the tangent of the angle that the line makes with the x axis has a value of m. When the straight line is at an angle of 45 degrees to the x axis, its tangent is 1 so m=1. If the line slopes backwards at 45 degrees to the x axis, it's tangent is -1 and m=-1. Forward sloping lines have a positive m, while backward sloping lines have a negative m. The value of b is also called the y intercept, because it's the point on the y axis where the straight line crosses that axis. It can have a positive or negative value. b is a constant, just like m. mx is m times x. The x axis is divided by equally spaced numbers, 0, 1, 2, 3 etc to the right, and -1, -2, -3 etc to the left of the origin. The y axis is similarly divided, postitive numbers going up and negative numbers going down. By putting numbers in the equation you can work out where points go on the line. m will have a value, like 2, for example, and b a value, say, 3, so we have y=2x+3. If we put x=0 we get y=3 which is the y intercept. So we mark that point (0,3) by going up 3 divisions on the y axis. Now put x=1, then y=5. So we move to the point (1,5), which is right 1 and up 5. We can join that point to (0,3) and continue beyond these points. What we find is that, although we have only plotted two points, other values of x and y actually fit on the line. If we look at where x=3 and go up to meet the line, then go horizontally back to the y axis, we should find it meets the point 9 on the y axis. So the line represents the relationship between x and y as given by the equation for all points including points in between our whole number divisions, like, for example, 1.5 or one and a half, halfway between 1 and 2.

### The equation of the circle is x^2+y^2+6x+2y-6=0 Find the exact values of k for which y=x+k is a tangent to the circle.

The equation of the circle can be written (x+3)^2+(y+1)^2=16 by completing squares for x and y. From this (y+1)^2=16-(x+3)^2. The centre of the circle is at (-3,-1) and the radius is 4. Differentiating wrt x we get: 2(x+3)+2(y+1)y'=0 so y'=-(x+3)/(y+1). This is the gradient at point (x,y), so since the gradient of the tangent has to be 1, because y=x+k has a gradient of 1, y'=1. If the point where the tangent line meets the circle is (p,q) then the tangent line is y-q=x-p so y=x+q-p and k=q-p. -(p+3)/√(16-(p+3)^2)=1, being the gradient at (p,q) or (p,√(16-(p+3)^2)). Therefore -(p+3)=√(16-(p+3)^2); squaring both sides we have: (p+3)^2=16-(p+3)^2 and (p+3)^2=8, p=-3±2√2. From the standardised equation of the circle (q+1)^2=16-8=8 and q=-1±2√2. There are two points on the circle where the tangent line has a slope of 1 so there are two possible values for k (see picture). These points are diametrically opposite. The first point is when x is negative and y is positive (2nd quadrant) and the opposite point is when x is negative and y is negative (3rd quadrant). The points are (-3-2√2,-1+2√2) and (-3+2√2,-1-2√2), giving values of k=2+4√2, 2-4√2. Note that the line y+1=x+3, that is, y=x+2 is the line passing through the circle's centre parallel to y=x+k. This line is equidistant from the two lines y=x+2+4√2 and y=x+2-4√2. From this fact it is possible to calculate the equations of the lines y=x+k for the two values of k without resorting to calculus. The two tangent points are radially distant from the centre by 4, at opposite ends of the diameter. The radius is the hypotenuse of an isosceles right-angled triangle, so the two sides have length a, given by 2a^2=4^2, making a=√8=2√2. The points on the circumference can then be calculated: y=-3±2√2 and x=-1±2√2. By plugging in the x-y values into y=x+k, or using slope-intercept form, the two values of k can be calculated: y+1-2√2=x+3+2√2, y=2+4√2; y+1+2√2=x+3-2√2, y=2-4√2.

### Identify the focus, directrix, and axis of symmetry of x=2y^2. Graph the equation.

This is the equation of a parabola, but it's lying on its side arms on either side of the x axis, which is the axis of symmetry, and the arms are on the positive side of x. The parabola's vertex is at the origin. The focus lies along the axis of symmetry and the directrix is a line x=-h. The rule about a parabola is that any point on the curve is equidistant from the directrix line and the focus point. Because of this rule when y=0, x=0 so the origin is midway between the directrix and focus, therefore the focus is at (h,0). Take any point P(x,y) on the curve. It's distance from the directrix is h+x; its distance from the focus is given by Pythagoras: sqrt((h-x)^2+y^2). But we know the equation of the parabola is x=y^2 so we can replace x and: h+2y^2=sqrt((h-2y^2)^2+y^2). Squaring both sides: (h+2y^2)^2=(h-2y^2)^2+y^2, y^2=(h+2y^2-h+2y^2)(h+2y^2+h-2y^2) (difference of two squares), y^2=8y^2h, h=1/8, because the y^2 cancel out. So the focus is at (1/8,0) and the directrix line is x=-1/8. You've got the general picture of how the parabola looks, so you can mark the focus and directrix line, you know the vertex is at (0,0), and you can plot a few points to help you get the curve right. When y=1 and -1 x=2 and when y=2 and -2 x=8. That gives you 4 points: (2,1), (2,-1), (8,2), (8,-2). See how the x axis acts like a mirror reflecting each half of the curve.

### Find the pts on the graph of z=xy^3+8y^-1 where the tangent plane is parallel to 2x+7y+2z=0.

Question: Find the pts on the graph of z=xy^3+8y^-1 where the tangent plane is parallel to 2x+7y+2z=0. The tangent plane to the surface z=f(x,y)=xy^3+8/y is given by 2x+7y+2z=D Where D is some constant. The equation of the tangent plane to the surface f(x,y), at the point (x_0 y_0 ) can be written as z=f(x_0,y_0 ) + f_x (x_0 y_0 )∙(x-x_0 ) + f_(y ) (x_0,y_0 )∙(y-y_0 ) Where (x_0,y_0) is some point on that plane andf_x (x_0 y_0 ) is the slope of the tangent line to the surface at that point in the x-direction and similarly for f_(y ) (x_0,y_0 ). We have f(x,y)=xy^3+8/y. f_x=y^3,  f_y=3xy^2-8/y^2 . So, f_x (x_0 y_0 )=y_0^3, f_(y ) (x_0,y_0 )=3x_0 y_0^2-8/(y_0^2 ), f(x_0,y_0 )=x_0 y_0^3+8/y_0 . Substituting for the above into the equation of the tangent plane, z = x_0 y_0^3+8/y_0 + y_0^3∙(x-x_0 ) + (3x_0 y_0^2 - 8/(y_0^2 ))∙(y-y_0 ) z = x_0 y_0^3+8/y_0 + y_0^3 x - x_0 y_0^3 + (3x_0 y_0^2 - 8/(y_0^2 ))y - 3x_0 y_0^3 + 8/y_0 z = y_0^3 x + (3x_0 y_0^2 - 8/(y_0^2 ))y + {x_0 y_0^3 + 8/y_0 - x_0 y_0^3 - 3x_0 y_0^3 + 8/y_0 } z = y_0^3 x + (3x_0 y_0^2 - 8/(y_0^2 ))y + {16/y_0 - 3x_0 y_0^3 } -2y_0^3 x - 2(3x_0 y_0^2 - 8/(y_0^2 ))y + 2z = 32/y_0 - 6x_0 y_0^3 We have just written down the equation of the tangent plane, which has also been written down as 2x+7y+2z=D Comparing the two forms of these equations, -2y_0^3 = 2     ----------------------------------- (1) 2(3x_0 y_0^2 - 8/(y_0^2 )) = 7     ---------- (2) 32/y_0 - 6x_0 y_0^3 = D   ------------------- (3) From (1), y_0 = -1 Substituting for y_0=-1 into (2), we get x_0 = 3/2. Substituting for x_0 and y_0 into (3), we get D = -23, and z = f(x_0,y_0 ) = (3/2) (-1)^3 + 8/(-1) = (-9.5) The equation of the tangent plane is thus 2x+7y+2z=-23 And this plane is tangential to the surface f(x,y) at one point only, which is: (1.5,-1,-9.5)

### Find the point on the curve y=x^2 which is nearest to the point A(0,3)

To find the nearest point on the curve we're looking for the line that is perpendicular to the tangent (normal). First we find the gradient by differentiation: 2x.  The normal has a gradient of -1/(2x) because the product of the slope of the normal and the slope of the tangent (gradient) is -1 and 2x*(-1/(2x))=-1. The gradient of the normal varies according to the position on the curve. The equation of the normal must pass through the point A. So y=mx+b must be satisfied by x=0 and y=3: so b=3 and the equation is y=mx+3, where m is to be determined. This line meets the curve when x^2=mx+3. If we put m=-1/(2x), we get x^2=-1/2+3=5/2 and x=+sqrt(2.5) and y=2.5. Therefore the nearest point is either (sqrt(2.5),2.5) or (-sqrt(2.5),2,5). Since the parabola is symmetrical the points are equidistant from A because A is on the axis of symmetry. The normal at (-sqrt(2.5),2.5) is the reflection of the normal at (sqrt(2.5),2.5). [The equations of the two normals are y=3-x/sqrt(10) (right-hand) and y=3+x/sqrt(10) (left-hand). The distance from A=sqrt(2.75).]

### Find the equation of the: a) tangent to the circle with center (-1,2) at the point (3,1) b) perpendicular bisector bisector of (AB) for A (2,6) and B(5,-2).

Problem: Find the equation of the: a) tangent to the circle with center (-1,2) at the point (3,1) b) perpendicular bisector bisector of (AB) for A (2,6) and B(5,-2). I  need help. a) Any line that is tangent to the circle has a slope that is    the negative inverse of the slope of the radius at the tangent    point.    For the radius: m = (y1 - y2)/(x1 - x2)    m = (2 - 1)/(-1 - 3)    m = 1/-4    m = -1/4    The slope of the tangent line is m = 4    The equation is found by using the slope and one point on    the line. We are given that one point: (3, 1)    y = mx + b    b = y - mx    b = 1 - 4(3)    b = 1 - 12    b = -11    Our equation is y = 4x - 11 b) As with the first part, we need the slope of the line we    were given. m = (y1 - y2)/(x1 - x2)    m = (6 - (-2))/(2 - 5)    m = (6 + 2)/(-3)    m = 8/-3    m = -8/3    The slope of any line perpendicular to this line is m = 3/8    The midpoint of the given line is ((x1 + x2)/2, (y1 + y2)/2)    (x1 + x2)/2 = (2 + 5)/2 = 7/2 = 3.5    (y1 + y2)/2) = (6 + (-2))/2 = 4/2 = 2    The midpoint is (3.5, 2)    We find the y-intercept with the formula b = y - mx    b = 2 - (3/8)3.5    b = 2 - 1.3125    b = 0.6875    Our equation is y = 3/8 x + 0.6875

### Tangent Line

1. y=ln(x); dy/dx=1/x; this is the tangent or gradient at point (x,ln(x)). Put in x=e and we get dy/dx=1/e and we get the point (e,1). The equation of the tangent is normally represented as a standard linear equation y=mx+c where m=dy/dx (at a particular point) and c is found by inserting the coords of the point of the tangent. So in the case in question, the point is (e,1) and dy/dx=1/e. y=x/e+c, 1=e/e+c so c=0 and y=x/e is the equation of the tangent line. 2. xy=5 (hyperbola) is differentiated thus: xdy/dx+y=0 (this uses the rule for differentiating a product uv, where u=x and v=y; d/dx(uv)=udv/dx+vdu/dx=xdy/dx+y). So dy/dx=-y/x=-(1/2)/10=-1/20. The equation of the tangent is in the form y=mx+c, where m=-1/20 and c is found by substituting the point coords: 1/2=-10/20+c; c=1/2+1/2=1, so y=1-x/20 or 20y=20-x is the equation of the tangent at (10,1/2).