Drawing two perpendicular tangent line from the origin to y=x2−2x+a.We drew two perpendicular tangent line from the origin (the point (0,0) ) to the curve y=x2−2x+a, what is the value of a?1) 542) −543) 344) −34

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SofZookywookeoybd · Accepted Answer

Let the tangent be y=mx, (as it passes through the origin), then we have the quadratic x2−(m+2)x+4−4a=0. In order for the given line be tangent to the given parabola, it must intersect at only one point, and hence B2−4AC=0). So we get a quadratic for m as m2+2m−4(1−a)=0. This means that two tangents are possible, and they will be perpendicular if m1m2=−1⇒4(1−a)=−1⇒a=54 (product of roots of a quadratic ax2+bx+c=ca)

Ishaan Stout · Accepted Answer

Let&amp;nbsp;f(x)=x2-2x+a.A line passing through (0,0) is of the form&amp;nbsp;y=λx. If such a line is tangent to the graph of f at some point&amp;nbsp;(α,β), then:1.&amp;nbsp;β=f(α)=α2−2α+a;2.&amp;nbsp;λ=f′(α)=2α−2.Therefore,&amp;nbsp;β=λα=2α2−2α&amp;nbsp;and&amp;nbsp;β=α2−2α+a,and therefore . So,&amp;nbsp;α=a&amp;nbsp;or&amp;nbsp;α=−a. In the first case,&amp;nbsp;β=2a−2a, and in the second case&amp;nbsp;β=2a+2a. In the first case, the slope of the tangent line will be&amp;nbsp;2a−2&amp;nbsp;and in the second case it will be&amp;nbsp;−2a−2. The only situation in which the tangent lines will be orthogonal, that is, the only case in which the product of these numbers is -1, is when&amp;nbsp;a=±54Hence,&amp;nbsp;a=54.

Drawing two perpendicular tangent line from the origin

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