Find equations of both lines through the point (2, -3) that are tangent to the parabola y = x squared + x.?
y = x² + x Take the derivative. The slope of the line will have to match the slope of the curve at the point of tangency. m = dy/dx = 2x + 1 Solve for x at the point of tangency. y + 3 = m(x - 2) x² + x + 3 = (2x + 1)(x - 2) x² + x + 3 = 2x² - 3x - 2 x² - 4x - 5 = 0 (x + 1)(x - 5) = 0 x = -1, 5 x² + x = y 1 - 1 = 0 25 + 5 = 0 The points of tangency are P(-1, 0) and (5, 30). The slopes of the tangent lines are: m1 = ∆y/∆x = (0 + 3) / (-1 - 2) = 3/(-3) = -1 m2 = ∆y/∆x = (30 + 3) / (5 - 2) = 33/3 = 11 The equations of the tangent lines are: y + 3 = -1(x - 2) y + 3 = 11(x - 2) or y = -x - 1 y = 11x - 25 ___________ If the point (2, 7) is inside the parabola, there is no tangent from the point to the parabola. Since the parabola opens upwards, this will be true if: y > x² + x 7 > 2² + 2 7 > 6 This is true so the point is inside the parabola so there is no tangent from the point to the parabola.
Find the Equations of all lines tangent to Y = x^2 -6......?
Hi, These tangent lines are y = 12x - 42 and y = 4x - 10. If f(x) = x² - 6, then f"(x) = 2x. This means that 2x is the slope of a tangent line when "x" is the coordinate of the point on the graph (x,x² - 6). Since slope = (y2 - y1)/(x2 - x1): ..........x² - 6 - 6 2x = -------------- ............x - 4 This simplifies to: 2x(x - 4) = x² - 12 2x² - 8x = x² - 12 x² - 8x + 12 = 0 (x - 6)(x - 2) = 0 x = 6 or x = 2 are the two x values on the graph of y = x² - 6 that have tangents going through (4,6). If x = 6, then the slope of 2x is 2(6) = 12. The equation in point-slope form would be y - 6 = 12(x - 4). y - 6 = 12(x - 4) y - 6 = 12x - 48 y = 12x - 42 <== FIRST TANGENT If x = 2, then the slope of 2x is 2(2) = 4. The equation in point-slope form would be y - 6 = 4(x - 4). y - 6 = 4(x - 4) y - 6 = 4x - 16 y = 4x - 10 <== SECOND TANGENT I hope that helps!! :-)
Find equations of both lines through the point (2,-3) that are tangent to the parabola y = x^2 + x?
The equation of a line having slope m and passing via way of (2, -3) is y + 3 = m(x - 2) ... ( a million ) fixing it with the equation of the parabola, m(x - 2) - 3 = x^2 + x => x^2 + (a million - m)x + 2m + 3 = 0 For the line to be the tangent to the parabola, the above equation in x will ought to have the comparable 2 roots => its discriminant = 0 => (a million - m)^2 - 8m - 12 = 0 => m^2 - 10m - 11 = 0 => (m -11) (m + a million) = 0 => m = 11 or - a million Plugging m = 11 and m = -a million in eqn. ( a million ) grants the two equations of the tangents as y + 3 = 11(x - 2) and y + 3 = - a million * (x - 2) => y = 11x - 25 and y = - x - a million For the given parabola, y - x^2 - x = 0, no tangent could be drawn from a element (x1, y1) that's interior the concave component to the parabola and the condition to verify that's y1 - x1^2 - x1 > 0 For the element (x1, y1) = (2, 7), the is 7 - 2^2 - 2 = a million > 0 and is confident using this that no tangent is often drawn from that factor to the parabola.
How do I find equations for the two tangent lines through the origin that are tangent to the curve [math]x^{2} - 4x + y^{2} + 3 = 0[/math]?
You know the tangent line goes through the origin and a point on the curve. So we need to find the two coordinates of the point on the curve (that's two unknowns). We have the equation of the curve. That's one equation. [math]x^{2} - 4x + y^{2} + 3 = 0[/math]We also know that the slope of the line will be the y-coordinate divided by the x-coordinate (because the line goes through the origin). So [math]y' = \frac y x[/math]. That's a second equation. But it has a knew unknown, [math]y'[/math].But using implicit differentiation, we know that [math]2x - 4 + 2y y'=0[/math]. That's a third equation.If we can find an [math](x,y)[/math] and a [math]y'[/math] that satisfy all three, we've found a point that does the job.Here's the algebra...Eliminating [math]y'[/math] by combining the second and third equations gives:[math]2x - 4 + 2y \frac y x=0[/math] Solving for [math]y^2[/math] gives:[math] y^2=2x-x^2[/math]Substituting this into the original equation for the curve gives:[math]x^{2} - 4x + 2x-x^2 + 3 = 0[/math][math] - 2x + 3 = 0[/math][math]x=1.5[/math]Then [math] y^2=2(1.5)-(1.5)^2 = 0.75[/math] so [math]y=\pm\frac{\sqrt{3}}2[/math].So we have a pair of potential solutions: [math]\left(\frac 3 2, \frac{\sqrt{3}}2\right)[/math] and [math]\left(\frac 3 2, -\frac{\sqrt{3}}2\right)[/math]These should be verified by checking that the slope at these points does match the ratio, [math]\frac y x[/math]. Both work.The slopes of the lines are [math]\pm \frac{\frac {\sqrt 3}2}{\frac 3 2}=\pm \frac {\sqrt 3}{3} = \pm \frac 1 {\sqrt 3}[/math]. Since they go through the origin, the equations for the lines are [math]y =\pm \frac x {\sqrt 3}[/math].Here's the picture:
Find equations of both lines through the point (2,-3) that are tangent to the parabola y = x^2+x.?
The equation of a line having slope m and passing through (2, -3) is y + 3 = m(x - 2) ... ( 1 ) Solving it with the equation of the parabola, m(x - 2) - 3 = x^2 + x => x^2 + (1 - m)x + 2m + 3 = 0 For the line to be the tangent to the parabola, the above equation in x should have the same two roots => its discriminant = 0 => (1 - m)^2 - 8m - 12 = 0 => m^2 - 10m - 11 = 0 => (m -11) (m + 1) = 0 => m = 11 or - 1 Plugging m = 11 and m = -1 in eqn. ( 1 ) gives the two equations of the tangents as y + 3 = 11(x - 2) and y + 3 = - 1 * (x - 2) => y = 11x - 25 and y = - x - 1 For the given parabola, y - x^2 - x = 0, no tangent can be drawn from a point (x1, y1) which is inside the concave part of the parabola and the condition to verify this is y1 - x1^2 - x1 > 0 For the point (x1, y1) = (2, 7), the condition is 7 - 2^2 - 2 = 1 > 0 and is satisfied which means that no tangent can be drawn from that point to the parabola.
Find equations of the tangent lines to the curve y = x − 1 /x + 1 that are parallel to the line x − 2y = 3.?
To find lines that are parallel to the line whose equation is: x - 2y = 3 You should write the equation in the slope-intercept y = mx + b: -2y = -x + 3 y = (1/2)x - 3/2 The slope of this line is 1/2. Now let's compute an equation that will give us the slope a tangent line to the given curve at any given point by computing the first derivative: dy(x)/dx = d[(x -1)/(x + 1)]/dx To compute this we will use the quotient rule f'(g(x)/h(x)) = {h(x)g'(x) - h'(x)g(x)}/{h(x)}² let g(x) = x - 1, then g'(x) = 1 let h(x) = x + 1, then h'(x) = 1 dy(x)/dx = f'(g(x)/h(x)) = {(x + 1)(1) - (1)(x - 1)}/{x + 1}² dy(x)/dx = {x + 1 - x + 1)}/{x + 1}² dy(x)/dx = 2/{x + 1}² Now we set the right-hand side equal to 1/2 and then solve for x: 2/{x + 1}² = 1/2 4 = {x + 1}² {x + 1} = ±2 x = 1 x = -3 y(1) = (1 - 1)/(1 + 1) = 0 y(-3) = (-3 - 1)/(-3 + 1) = -4/-2 = 2 The points where the slope (m = 1/2) are (1, 0) and (-3, 2) Using the point-slope for of a line, y - y1 = m(x - x1) we can write the equations for these two lines: y - 0 = (1/2)(x - 1) y - 2 = (1/2)(x - (-3)) y = (1/2)(x - 1) y = (1/2)(x + 3)) + 2 y = (1/2)x - 1/2 y = (1/2)x + 7/2 The y -intercepts are -1/2 and + 7/2
Find the equations of all tangent lines to the graph of f(x) = 4x- x^2 that pass through (2,5)?
y = 4x- x^2 dy/dx = -2x + 4 therefore x= 2 is a maximum (2,4) is the maximum From (2,5) there will be two tangents, one slanting to the right and one slanting to the left, crossing at (2,5). OK, so let a be any particular value of x. So a general Point on the parabola is (a, (4a-a²)) The Slope at x=a is (4-2a) We are looking for two special values of a, one will be greater than 2 and the other one will be less than 2. You can think the one where a is greater than 2. So for that point, the slope is negative. Draw a picture. So the tangent at point (a, (4a-a²)) crosses the line x=2 at some point where y is greater than (4a-a²) How much greater? Slope times (2-a) [this is the hardest step] So tangent goes through the point (2, ((4a-a²)+(2-a)(4-2a)) Simplify the y coordinate is (4a- a²)+(8+2 a²-8a) y coordinate is -4a+ a²+8 Set that = 5, and you know the value of a that satisfies the condition that the tangent goes through (2,5) Now you have two values of a. They should be equidistant from x=2. So you also have two values for the slope. They should be equal and opposite. one point (2,5), and 2 values for slope gives you two lines that intersect at (2,5) and are both tangent to the parabola. Check my calculations, but that is the approach. You will have two values of a because you solved a quadratic.
How do you find the equation of the tangent line for the function f(x) = 5x^2 at the point (2,20)?
what's the curve equation to discover the tangent line at a factor (25,5) on it? EDIT: [As in retaining with the greater suitable information, which i'd desire to relook in complication-free words after approximately 9 hours of earlier presentation] a million) Differentiating the given one, dy/dx = a million/(2?x) 2) At x = 25, dy/dx = a million/(2?25) = a million/10; this is the slope of the tangent line on the given factor; this is by using the geometrical definition of differentiation] 3) using Slope-factor sort, the equation of the tangent line at (25,5) is: y - 5 = (a million/10)(x - 25) increasing and simplifying, the equation is: x - 10y + 25 = 0
What is the answer to find an equation of the tangent line to the graph of f(x) =2x^2-7 at the point (3,11)?
Given, f(x)=2x^2-7=>f'(x)=4xwhich implies that the slope at x=3 is 4*3=12The tangent is of the form y=mx+c, where m is the slope and c is the y-intercept.We know m=12, substituting in the equation,y=12x+cSubstituting (x,y) as (3,11) in this equation, 11=12*3+c => c=11-36=>c=-25.Therefore, the equation of the tangent is y=12x-25.
What is the equation of the tangent line from the point [math]P(0,4)[/math] to the curve [math]y=\sqrt {9-x^2}[/math]?
In its mathematical heft this problem, hands down, is nowhere near even the simplest of Olympic problems (of all times) but in spirit, if you choose to take that stance, is very much so:find a well-disguised elementary solution (to a difficult problem)Before you read on any further stop and ponder - do you see one?Reason without paper. Just in your mind’s eye.I upvoted the answer by Doug Dillon and the credit for what follows goes to him.So as I was gazing at Mr. Dillon’s numbers I was thinking:- … hm … square … root … of … seven … huh … over .. three …As Alon aptly puts it, I was hearing the music but wasn’t quite sure what the tune was.I can explain three. That’s the radius. But the square root of seven? Sure, it’s a nice prime number. But where did it come from …Aaaaaaaa …. I see now. A rectangle on the secant is a square on the tangent, a computationally cheap theorem otherwise known as Euclid’s Book 3 Proposition 36:[math]1 \cdot (1 + 3 + 3) = t^2 \tag{1}[/math][math]t = \pm \sqrt{7} \tag{2}[/math]Double check? By the slightly more expensive Pythagorean theorem:[math](1 + 3)^2 - 3^2 = 7 \tag{3}[/math]Bisect the line segment [math]OP[/math] with [math]b[/math] to locate the point of their intersection on the [math]y[/math]-axis [math](0, 2)[/math]:Construct a circle [math]s[/math] with the center at [math](0, 2)[/math] and the radius [math]2[/math] to locate the points [math]A[/math] and [math]B[/math]:By the Thales’ Theorem also known as Euclid’s Book 3 Proposition 31 the angle at [math]A[/math], that subtends the diameter [math]OP[/math], is right - just what we need for a tangent.Next comes the step popular in problem solving - introduce an auxiliary item that was not originally present but is helpful - complete the circle, shown in green, to set up the B3P36 theorem:The secant now is [math]PD[/math] such that [math]|PD| = 7[/math]. Together with [math]|PC| = 1[/math] by B3P36 we have:[math]|PD|\cdot |PC| = |PA|^2 \tag*{}[/math][math]|PA| = \sqrt{7} \tag*{}[/math](also via Pythagoras from [math]\triangle PAO[/math]).Lastly, the right, by construction, triangles [math]PAO[/math] and [math]POX_a[/math] are similar, the tangent, as a trigonometric function, of the angle [math]\theta[/math], by definition from [math]\triangle PAO[/math], is [math]\sqrt{7}[/math] over [math]3[/math], the [math]y[/math]-intercept is [math]4[/math] and the equations of the tangent straight lines sought-after are:[math]y = \pm \dfrac{\sqrt{7}}{3}x + 4 \tag*{}[/math]