##### parabola of a tangent

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For what values of a and b is the line

4x + y = b
tangent to the parabola
y = ax2
when x = 4?

Jul 15th, 2015

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$y=ax^2$

Differentiating y w.r.t. x

$\\ \frac{dy}{dx}=\frac{d}{dx}(ax^2)\\ \\ \frac{dy}{dx}=2ax\\$

This is the slope of tangent to the parabola at any point x.

At x = 4

Slope of the tangent is

$\frac{dy}{dx}=2a\times4=8a$

Given that

4x + y = b

is a tangent to the parabola at x = 4

The equation of tangent can be rewritten as

y = -4x + b

Comparing the above equation with slope intercept form of a line

y = mx + b (where m is the slope of the line)

we get, slope of the tangent = -4

Therefore,

$\\ 8a=-4\\ \\ a=\frac{-4}{8}\\ \\ a=-\frac{1}{2}$

So, equation of parabola becomes

$y=-\frac{x^2}{2}$

At x = 4

y = $-\frac{4^2}{2}=-\frac{16}{2}=-8$

(4, -8) is the point at which the tangent touches the parabola.

Substitute x = 4 and y = -8 in the equation of tangent to find b

$\\ 4\times4+(-8)=b\\ \\ 16-8=b\\ \\ 8=b\\ \\ b=8$

ANSWER:  $a=-\frac{1}{2},\hspace{5}b=8$

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Jul 15th, 2015

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Jul 15th, 2015
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Jul 15th, 2015
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