show y=3x-5 is tangent to x^2 + y^2 +2x -4y - 5 = 0 and the point where they touch

y=3x-5x^2 + (3x-5)^2 + 2x - 4(3x-5) - 5 = 0x^2 + 9x^2 -30x +25 + 2x -12x + 20 - 5 = 010x^2 -40x + 40 = 010 (x^2 - 4x +4) = 010(x - 2)^2 = 0x=2implies one point of contact, therefore tangenty = 3x - 5y = 6 -5 = 1

RM

Related Maths Scottish Highers answers

All answers ▸

The equation x^2 + (k-5)x + 1 = 0 has equal roots. Determine the possible values of k.


Express '2x^2 + 8x + 30' in the form 'a(x+b)^2 + c'


Calculate the rate of change of y(t) = 1/(4t), when t = 8


log_a(36) - log_a(4) = 0.5, what is a?