 Javion Kerr

2022-04-05

Drawing two perpendicular tangent line from the origin to $y={x}^{2}-2x+a$.
We drew two perpendicular tangent line from the origin (the point (0,0) ) to the curve $y={x}^{2}-2x+a$, what is the value of a?
1) $\frac{5}{4}$
2) $\frac{-5}{4}$
3) $\frac{3}{4}$
4) $\frac{-3}{4}$ SofZookywookeoybd

Let the tangent be $y=mx$, (as it passes through the origin), then we have the quadratic ${x}^{2}-\left(m+2\right)x+4-4a=0.$ In order for the given line be tangent to the given parabola, it must intersect at only one point, and hence ${B}^{2}-4AC=0\right).$ So we get a quadratic for m as ${m}^{2}+2m-4\left(1-a\right)=0$. This means that two tangents are possible, and they will be perpendicular if ${m}_{1}{m}_{2}=-1⇒4\left(1-a\right)=-1⇒a=\frac{5}{4}$ (product of roots of a quadratic $a{x}^{2}+bx+c=\frac{c}{a}\right)$ Ishaan Stout

Let $f\left(x\right)={x}^{2}-2x+a$.
A line passing through (0,0) is of the form $y=\lambda x$. If such a line is tangent to the graph of f at some point $\left(\alpha ,\beta \right)$, then:
1. $\beta =f\left(\alpha \right)={\alpha }^{2}-2\alpha +a$;
2. $\lambda ={f}^{\prime }\left(\alpha \right)=2\alpha -2$.
Therefore,
and therefore . So, $\alpha =\sqrt{a}$ or $\alpha =-\sqrt{a}$. In the first case, $\beta =2a-2\sqrt{a}$, and in the second case $\beta =2a+2\sqrt{a}$. In the first case, the slope of the tangent line will be $2\sqrt{a}-2$ and in the second case it will be $-2\sqrt{a}-2$. 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 $\sqrt{a}=±\sqrt{\frac{5}{4}}$
Hence, $a=\frac{5}{4}$.

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