$r$ is tangent at two points on the parabola $y=x^2$ such that the angle between the two radii at the tangent points is $2\theta$, where $0<2\theta<\pi$. Find $r$ as a function of $\theta$.
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- What is the slope of the tangent to the curve
$y = 2x^2+3x-2$at$x=4$? - Find the value of
$a$such that the line$y_1=3x+a$is tangent to the curve$y_2=2x^2+3x+1$ - You are given
$3$lines:$y=2x$,$y=2x-2$,$y=-\frac{1}{2}x+3$. For each pair of lines, state whether they are parallel or perpendicular.
- What is the slope of the tangent to the curve
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Let
$P$be one of the points where the parabola and the circle touch. What is the slope of the tangent at$P$, given you know the equation of the curve? -
What is the angle made by the tangent at
$P$with the$x$-axis in terms of$\theta$? -
What is another expression of the slope of the tangent at
$P$in terms of this angle. -
Can you extract the tangent of this angle from the right triangle with sides
$r,$$x$and$r\cos(\theta)?$
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Let $P=(x,x^2)$be one of the two points where the parabola and circle are tangent to each other. The radius$r$sits on the line normal at$P$. The slope of the tangent line at$P$is equal to the derivative of$x^2$, i.e.$2x$, but also equal to the tangent of the angle the line makes with the$x$-axis, which in this case is$\theta$(owing to the fact we have a right angle at$P$between the tangent and the normal), i.e.$\tan\theta=\frac{x}{r\cos\theta}$. Equating we get$2x=\frac{x}{r\cos\theta}$and so$r=\frac{1}{2\cos\theta}$.Alternatively, we could also have used the fact that the product between the slopes of the tangent and the normal is
$-1$. The slope of the normal in our case is minus the tangent of the angle opposite$\theta$(can you explain why?), i.e.$-\frac{r\cos\theta}{x}$. Hence$r$is extracted from$-\frac{r\cos\theta}{x}\cdot2x=-1$.