Chapter 8.1, Problem 74E

Solution

To Show: A line tangent to the given parabola at the point $\left(a,b\right)$ crosses the $y$ axis at $\left(0,-b\right)$ .

It has been shown that the line tangent to the given parabola at the point $\left(a,b\right)$ crosses the $y$ axis at $\left(0,-b\right)$ .

Given:

The parabola, $x^2=4py$ .

Concept used:

The slope of the tangent line to any curve $y=f\left(x\right)$ at any point, is the value of $\frac{dy}{dx}$ at that point.

Calculation:

The given parabola is $x^2=4py$ .

Simplifying,

  $y=\frac{1}{4p}{x}^2$

Differentiating,

  $\frac{dy}{dx}=\frac{1}{2p}x$

Put $\left(x,y\right)=\left(a,b\right)$ in $\frac{dy}{dx}=\frac{1}{2p}x$ to get,

  $\frac{dy}{dx}\left(a,b\right)=\frac{a}{2p}$

So, the slope of the tangent line to the given parabola at the point $\left(a,b\right)$ , is $\frac{a}{2p}$ .

Now, clearly, the tangent line also passes through the point $\left(a,b\right)$ .

Then, the equation of the tangent, using point-slope form is:

  $y-b=\frac{a}{2p}\left(x-a\right)$

Now, put $x=0$ in the above equation to get,

  $y=b-\frac{a^2}{2p}$

Now, the point $\left(a,b\right)$ lies on the parabola $x^2=4py$ .

Then, $\left(a,b\right)$ must satisfy the equation, $x^2=4py$ .

Put $\left(x,y\right)=\left(a,b\right)$ in $x^2=4py$ to get,

  $a^2=4pb$

Put $a^2=4pb$ in $y=b-\frac{a^2}{2p}$ to get,

  $y=-b$

So, putting $x=0$ in the equation of the tangent; $y-b=\frac{a}{2p}\left(x-a\right)$ , gives $y=-b$ .

This shows that the line tangent to the given parabola at the point $\left(a,b\right)$ crosses the $y$ axis at $\left(0,-b\right)$ .

Conclusion:

It has been shown that the line tangent to the given parabola at the point $\left(a,b\right)$ crosses the $y$ axis at $\left(0,-b\right)$ .