# Find tangent line of curve that intersects point.

How do I find the tangent line of the curve $y=x^2$ that intersects the point $(8,2)$?

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What does the equation of a generic tangent line of your curve look like? –  Guess who it is. Apr 26 '12 at 2:18

Let the point of tangency be $(a,a^2)$. Then the slope of the tangent line is $2a$, and the equation of the tangent line is $$y-a^2=2a(x-a).$$ Substitute $x=8$, $y=2$, and solve the resulting quadratic for $a$.

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Starting with Andre's equation of $y-a^2=2a(x-a)$, $y=2ax-2a^2+a^2 = 2ax-a^2$, or $a^2-2ax = -y$ or $a^2-2ax + x^2 = x^2-y$ or $(a-x)^2 = x^2-y$ or $a = x \pm \sqrt{x^2-y}$.

Note that this requires $x^2 \ge y$, which means that the point that the tangent goes through has to be below (or on) the parabola and that, if the point is below, there are two tangents going through it - see if you can visualize why.

Also note that the two values of $a$ are equally spaced around $x$.

Also$^2$ note that if $y < 0$ then there are always two values of $a$, one of which is negative.

If $x=8$ and $y=2$, $x^2-y = 62$, so $a = 8 \pm \sqrt{62}$.

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Find $\frac{dy}{dx}$(derivative with respect of x) and substitute the point$(8,2)$ then you will get $\frac{dy}{dx}$ at $(8,2)$ which is slope of tangent line and you have slope and point$(8,2)$ then find equation using formula y - y1 = m(x - x1)

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It is y-16x+64=0

y(x)=x^2 y'(x)=2x

y(8)=64 y'(8)=16

Tangent Equation: y-y1=m(x-x1) y-64=16(x-8)

y-16x+64=0

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