Y= 4x^2 5-x /x X Does Not = 0 Show That Dy/dx At P Is 3 Find An Equation Of The Tangent To C

How do I find the equation of the tangent to the hyperbola, 12x^2 -4y^2= 48 at (5,3)?

I believe you have written something wrong in your question.You cannot draw a tangent from (5, 3) to this hyperbola.It needs to be a point on the curve. Perhaps you meant from (4, 6)?If that is so, I may as well do it…At (4, 6) the gradient is 2The equation of the tangent is of the form y = mx + c soy = 2 x + cand we find c by substituting x = 4, y = 66 = 8 + c so c = – 2The tangent is y = 2x – 2and here is the graph…

How do I determine the equation of the tangent to the curve [math] y=2x-x^2 [/math] that passes through point [math] (2,9) [/math]?

There are actually two such tangents:Let y = mx + n be the tangent we're looking for, i.e. we have to determine m and n.By definition of tangent, the line y = mx + n is a tangent to the given curve if they interesect in exactly one point. So let's equate the two equations to calculate intersection(s):[math]y = mx + n = 2x - x^2,[/math][math]x^2 + (m-2)x + n = 0,[/math][math]x = \frac{m-2}{2}\pm\sqrt{\left(\frac{m-2}{2}\right)^2-n}.[/math]If the radicand (the "thing" under the square root) is positive, there will be two intersections. If it is negative, there will be no intersection. If it is zero, there will be one intersection, and the line will indeed be a tangent.So we get:[math]\left(\frac{m-2}{2}\right)^2-n=0,[/math][math]n=\left(\frac{m-2}{2}\right)^2.[/math]Now we can use the fact that the tangent passes through (2,9):[math]9=2m+n=2m+\left(\frac{m-2}{2}\right)^2=2m+\frac{\left(m-2\right)^2}{4},[/math][math]36=8m+4-4m+m^2,[/math][math]m^2+4m-32=0,[/math][math]m=-2\pm\sqrt{4+32}=-2\pm\sqrt{36}=-2\pm 6,[/math][math]m=4\ \text{or}\ m=-8.[/math]For m = 4, we get[math]n=\frac{(2-4)^2}{4}=\frac{4}{4}=1,[/math]and for m = -8, we get[math]n=\frac{(2-(-8))^2}{4}=\frac{100}{4}=25.[/math]The two solutions are therefore y = 4x + 1 and y = -8x + 25.

How can one find the values of k for which the line 2x -k is tangent to the circle with the equation x^2 + y^2 = 5?

The simple way to do this is to clearly define what  it means for tangent so that finding the k values is the easiest. Problem Specific AnswerWe have [math]y = 2x - k[/math]  and [math]x^2 + y^2 = 5[/math] and the line is tangent to the circle. What it means for a 2 things to be tangent is that one point and only one point satisfies both equations at the same time. That means that in the above two equations,  only the point represented by  [math](x,y)[/math] satisfies both equations. We get:  [math]y = 2x - k[/math] and [math]x^2 + y^2 = 5 [/math]Substituting, we get [math]x^2 + (2x-k)^2 = 5.[/math] This equation must only have one solution. [math]5x^2 -4kx +(k^2 - 5) = 0.[/math]This quadratic equation must have its determinant equal to 0 in order for the two to be tangent. [math]b^2 - 4ac = 0[/math][math]16k^2 - 20k^2+100 = 0[/math][math]4k^2 = 100[/math][math]\boxed{k = \pm 5}[/math]Generic AnswerWe can generalize this solution for any line and circle.Line: [math]y = mx+b[/math]Circle: [math]x^2 + y^2 = r^2[/math]Substituting we get: [math]x^2 + (mx+b)^2 = r^2[/math][math]\Rightarrow (m^2 + 1)x^2 + 2mbx + (b^2 - r^2) = 0[/math]Taking the discriminant and making it equal to 0, we get[math]4m^2b^2 - 4(m^2 + 1)(b^2 - r^2) = 0[/math][math]4m^2b^2 - 4m^2b^2 + 4m^2r^2 - 4b^2 + 4r^2 = 0[/math][math]m^2r^2 + r^2 = b^2[/math][math]m, r,[/math] and [math]b[/math] must satisfy these equations in order to be tangent. We can check out previous answer by plugging in here: [math]m = 2; r = \sqrt{5}; b = -k.[/math][math]20 + 5 = k^2.[/math]Therefore, [math]\boxed{k = \pm 5}[/math]

How can I find all the points at which the tangent passes through the origin, for the curve [math]y=4x^3 - 2x^5[/math]?

Slope of tangent at point(h,k) = 12(h^2)-10(h^4)(h,k) lie on curve=> k = 4(h^3)-2(h^5)Tangent at this point =>(y - k) = m (x-h)(y - 4(h^3) + 2(h^5)) = (12(h^2)-10(h^4)) (x - h)Passes through origin.Put x=0,y=0Solving this, we get h=0,1,-1Thus the points are (0,0), (1,2), (-1,-2)

Does the parabola y=2x^2-13x+5 have a tangent whose slope is -1? If so, find the equation for the line and point of tangency. if no, why not?

dy/dx= 4.x-13at x=3, dy/dx=-1y= 2.3^2–13.3+5=-16so the tangent line with slope=-1 will tangent to parabola at (3, -16)equation of this tangent line is y=-x +mreplace the coordinate of the point of tangent in above equation we get m= -13so the equation of tangent line is y= -x-13