text: Part 1\nThe tangent at a point (a, b) to the hyperbola\n\nx²/16 - y²/4 = 1\n\nhas the equation \n\nax/16 - by/4 = 1 \n\n(similarly for other 2 deg curves). You can see why this is a straight line passing thru (a, b). It's less obvious that this is the tangent line (to show this, work out the gradient vectors of x²/16 - y²/4 and of ax/16 - by/4 at (a, b); they are the same up to scale, which means that the hyperbola and the straight line are perpendicular to the same vector, and therefore are tangent to each other).\n\nAt the point a = 5 and b = -3/2, the line's slope is a/4b = -5/6.\n\n\nIn Part 2, the "0" must mean the angle theta. \nIn polar coordinates, \n\nx = r cos(theta)\ny = r sin(theta)\n\nSo the equation y = 3x + 5 becomes \n\nr sin(theta) = 3r cos(theta) + 5\n\nor \n\nr = 5/[sin(theta) - 3cos(theta)]\n\n(which is C).\n\nYou can also use trig identities to rewrite this in the form r = D/cos(theta - a constant), which is the std form of the polar equation of a straight line (see, for example, http://www.xtec.es/~rgonzal1/treatise.pdf, page 41).