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Seems like this question should be straight forward, so I'm guessing it is the interpretation of the question that is messing me up.


Find the coordinates of all points on the graph of $y=1-x^2$ at which the tangent line passes thru $(2,0).$


My approach:

The tangent line to $f(x)$ has a slope of $-2x$. If it passes thru $(2,0)$ then I can write the equation of the tangent line as $y=-4x+8.$ So, I just need to find the intersections of this line and the parabola function given.

Setting the 2 y values equal to find the intersections, I get,

$-4x+8=-x^2+1$ or $x^2-4x+7=0$. This quadratic has imaginary roots, so something is not right in my logic or approach or method of solution. I've checked the arithmetic. Can you assist?

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  • $\begingroup$ The equation you wrote is not of the tangent line. $\endgroup$ – hamam_Abdallah Sep 12 '18 at 20:11
  • $\begingroup$ $(2,0)$ does not lie on the parabola. $\endgroup$ – amd Sep 12 '18 at 23:50
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Suppose the tangent touches the parabola at the point $P(x_0,1-x_0^2)$

The slope at the point $P$ $$m=-2x_0$$

The equation of the line passing through $(2,0)$ and with slope $m$ is

$$y=-2x_0(x-2)$$

But this line also passes through the point $P$

Thus $$1-x_0^2=-2x_0(x_0-2)$$

$$x_0=2\pm \sqrt 3$$ $$y_0=1-x_0^2=-6\mp 4\sqrt 3$$

Points are $P_1(2+\sqrt3,-6-4\sqrt 3)$ and $P_2(2-\sqrt 3,-6+4\sqrt 3)$

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  • $\begingroup$ totally clear now. thank you. $\endgroup$ – user163862 Sep 14 '18 at 19:01
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Your $x$ is doing double duty. It can't travel about the line and parabola and also be part of the slope formula at a certain point.

Instead, suppose the intersection happens at $x=a$. Then the equation of the line is $y=-2ax+4a$. At the intersection point you have

$$1-a^2 = -2a^2+4a$$

So $a=2\pm \sqrt{3}.$

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The Tangent line equation at $(a,1-a^2)$ is

$$y=-2a(x-a)+(1-a^2)$$

it goes through $(2,0)$ if

$$0=-2a(2-a)+(1-a^2)$$

or

$$a^2-4a+1=0$$ $$\Delta=12$$ thus

$$a=2\pm \sqrt{3}$$

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    $\begingroup$ How is $\Delta'$ defined? $\endgroup$ – N. F. Taussig Sep 13 '18 at 8:11
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    $\begingroup$ @N.F.Taussig $\Delta'=(\frac b2)^2-ac$ $\endgroup$ – hamam_Abdallah Sep 13 '18 at 19:29
  • $\begingroup$ Thanks for clarifying. $\endgroup$ – N. F. Taussig Sep 14 '18 at 0:00
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Let $P (t,1-t^2)$ be a point on the parabola .

Tangent line has slope $-2t$ at $P(t,1-t^2)$. It passes through $(2,0)$.

Slope of line joining $P$ and $(2,0)$ equals $-2t$.

$\dfrac{1-t^2}{t-2} = -2t.$

$t^2-4t+1=0.$

Solve for $t$.

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