Prove that the straight line y = mx + c touches the parabola y2=4a(x+a) if c=ma+am.
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Solution
satisfy the given line with parabola and then put D=b2−4ac=0 quad eqn.ax2+bx+c parabola y2=4a(x+a) line y=mx+c (mx+c)2=4a(x+a) (mx)2+2mxc+c2=4a(x+a) (mx)2+2mxc+c2−4ax−4a2=0 Now put D = 0 D=b2−4ac=0 (2mc−4a)2−4m2(c2−4a2)=0 a2−mca+m2a2=0 a2+m2a2=mca c=am+ma