calculusconic sectionstangent line
The books say's
"Find the equations of the tangents to $x^2+3y^2=4$ which are
perpendicular to the line $x-2y=7$"
I've graphed them and found that the given line does not pass through the ellipse and the gradient of the tangent should be $dy/dx=-x/3y$. I don't know how to get an expression for the tangent because all the examples we did had a given point as well or the line intercepted the curve and i could find a point using simEqu. please help!
We want the tangent line(s) to pass through $(27,3)$. As you observed, this point is not on the ellipse $x^2+9y^2=81$. That's perfectly all right, but it does make the problem harder. The point $(27,3)$ is outside the ellipse, so an informal picture shows there should be two lines through $(27,3)$ that are tangent to the ellipse.
Let the point of tangency be $(a,b)$. This point is on the ellipse, and therefore $$a^2+9b^2=81.\tag{$1$}$$
By implicit differentiation, $2x+18y y'=0$. It follows that the tangent line at $(a,b)$ has slope $-a/9b$.
The equation of the line through $(a,b)$ with slope $-a/9b$ is $$y-b=-\frac{a}{9b}(x-a).\tag{$2$}$$ The point $(27,3)$ is on this line. It follows that $$27-b=-\frac{a}{9b}(3-a).\tag{$3$}$$ Simplify Equation $(3)$, and use Equation $(1)$ to find the values of $a$ and $b$. Once you have found these, you will know the equations of the required lines.
$$2y+x+3=0$$ The important thing is that you know what being perpendicular means. In terms of slopes, one line has to have a slope that's equal to the negative reciprocal of the other. This is a property you should know from your Algebra class. For this problem, we want to find the slope of the line that's perpendicular to the line we have above. To do that, we will first need to find the slope of the line above. Let's rearrange stuff to find it: $$2y=-x-3 \\ y=-\frac12x-1.5$$
You can get $\frac{dy}{dx}$ of this, but really it's in the form of $y=mx+b$ and we know that the slope of this line is $-\frac12$. Thus, the slope of the line perpendicular to this one is $2$. $$4x^2+y^2=72$$ We've got the slope, but not the line yet. We want to find a line that will be tangent to the ellipse that was given. How do we find that? Start with finding the point at which it is tangent. You probably already know how to do this. First take the derivative. For this problem, implicit differentiation is not a really good idea. Thus, isolate for $y$ first and then solve it:
$$y=\sqrt{72-4x^2}$$
$$\frac{dy}{dx}=\frac12 \left(72-4x^2\right)^{-\frac12}(-8x)$$
Now that you have the derivative, find the value for $x$ where the slope of the ellipse is the same as the perpendicular line. In other words, $\frac{dy}{dx}=2$:
$$2=\frac12 \left(72-4x^2\right)^{-\frac12}(-8x)$$
A little messy, but you'll find that $x=\pm3$. Therefore the point on the ellipse where the tangent will have this slope is $(\pm3, \pm6)$ (plugged in to the original to get the $y-$value). From this point, you can easily get the line now. You've got $y=mx+b$. $m$ is $2$, and now to find $b$ we'll plug in $x/y$:
$$6=(2)(-3)+b$$
$$b=12$$
Or for $x=3$, $b=-12$.
Our final answer ends up being:
$$\therefore y=2x\pm12$$
Pictorial proof:
Best Answer
Hint: Slope of the equation $x-2y=7$ is $0.5$. So the slope of the tangent to the ellipse is $-2$. So we get $$-2=\dfrac{-x}{3y}$$ Then find the points at which the tangent cuts the ellipse using this equation and then get the equation(s).