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)$\frac{-3}{4}$

We drew two perpendicular tangent line from the origin (the point (0,0) ) to the curve

1)

2)

3)

4)

SofZookywookeoybd

Beginner2022-04-06Added 15 answers

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

Ishaan Stout

Beginner2022-04-07Added 14 answers

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 $(\alpha ,\beta )$, then:

1. $\beta =f\left(\alpha \right)={\alpha}^{2}-2\alpha +a$;

2. $\lambda ={f}^{\prime}\left(\alpha \right)=2\alpha -2$.

Therefore, $\beta =\lambda \alpha =2{\alpha}^{2}-2\alpha {\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\beta ={\alpha}^{2}-2\alpha +a,$

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 $\sqrt{a}=\pm \sqrt{\frac{5}{4}}$

Hence, $a=\frac{5}{4}$.

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A cap of a sphere with radius r and height h.

V=??

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b) $f\left(x\right)=-3{x}^{2}+7$

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?

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C)Kite;

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