Web(1) Equation of the line is y = mx + c ---- (2) Solving above equations, x2 = 4a (mx + c ) ⇒ x2 - 4amx -4a c =0 which is a quadratic in x. If the given line is a tangent to the parabola, the roots of above equation are real and equal. ⇒ b2 -4ac = 0 ⇒ 16a2 m2 +16ac =0 ⇒am2 +c =0 ⇒c = - am2 is the required condition. ← Prev Question Next Question →

Solve y=mx+c= Microsoft Math Solver

WebIf the line y = mx + c is a common tangent to the hyperbola -x2/100 - y2/64 = 1 and the circle x2 + y2 = 36, then which one of the following is true? (1) 5m = 4 (2) 4c2 = 369 (3) … WebAnswer (1 of 24): Point of intersection of the line x-2y =—1 and y =2 is (3 , 2). Line y = m.x +c passes through the point (3 , 2) , then 2 = 3.m + c ... glassdoor newson health

Intersection of a Line and Ellipse eMathZone

Web30 okt. 2011 · Since you didn't specify the types of the pair's x and y, my guess is that the line double m = (b.y-a.y)/ (b.x-a.x); is causing the problem. If the two values are int, and the first is lower, it will always return 0. For m to be the actual value, you need an explicit cast to bool: double m = (double) (b.y-a.y)/ (b.x-a.x); Share Improve this answer Web30 jan. 2013 · I was wondering if it is possible to plot a line of the form y = mx+b in Matlab? I used polyfit to get a 1x2 array that contains the slope and intercept. Here is what I have … Web7 dec. 2013 · How can I write for any arbitrary data set of x and y, first fit as y=mx+c and from this fit find c and then with the help of other line y=x it will find angle between two lines ? sixwwwwww on 7 Dec 2013 glassdoor newswhip