How to calculate coordinates of point P on an ellipse that has a tangent line at 45 degrees

Question

I have the equation of an ellipse: $\frac{x^2}{4}+\frac{y^2}{9}=1$

In the upper left quadrant, how do I find the coordinates of tangent point P whose tangent line is at an angle of 45 degrees with the x-axis (slope of 1)?

I haven't done this kind of math in nearly a decade, so I haven't been able to figure it out.

Answer 1

A way to do this without derivative:

Let the tangent line be $y=x+c$. Substitute into equation of the ellipse: $$\frac{x^2}{4}+\frac{(x+c)^2}{9}=1$$ $$\implies13x^2+8cx+(4c^2-36)=0\tag{*}$$

Since the tangent touches the ellipse (at exactly one point), the discriminant of $(*)$ has to be zero: $$\Delta=-144(c^2-13)=0\implies c^2=13\implies c=\pm\sqrt{13}.$$ $c=\sqrt{13}$ is obvious.

Back to $(*)$, $13x^2+8\sqrt{13}x+(4\sqrt{13}^2-36)=0\implies x=\dfrac{-4}{\sqrt{13}}$, a repeated root.

Answer 2

Find the derivative:

$$\frac{x}{2}+\frac{2yy'}{9}=0 \qquad \qquad(1)$$

You know that at some point on the ellipse $(x_1,y_1)$ the slope of the tangent is $y_1'=\tan 45^o = 1$

If you replace $x_1,y_1$ and $y_1'$ into the equation of ellipse and (1), you get the followwing system of two equations and two variables:

$$\frac{x_1^2}{4}+\frac{y_1^2}{9}=1 $$

$$\frac{x_1}{2}+\frac{2y_1}{9}=0$$

You can easily solve this system by eliminating one variable from the second equation and replacing it into the first. You'll get a quadratic equation with two solutions. Just pick the one from the second quadrant $(x_1\lt0)$

Answer 3

The ellipse has the parametric equation $x=2\cos t, y=3\sin t$ (with $t\in[0,2\pi])$. The tangent vector at $t$ is $(-2\sin t,3\cos t)$. It makes an angle $45^\circ$ with the horizontal if $-2\sin t=3\cos t$, that is $\tan t=-\frac{3}{2}$. Given that you want the point to the in the upper left quadrant, that means $t=\pi-\arctan\frac32$. Then the point on the ellipse has coordinates

$$x=2\cos(\pi-\arctan\frac32)=-2\cos\arctan\frac32$$ $$y=3\sin(\pi-\arctan\frac32)=3\sin\arctan\frac32$$

Let $\alpha=\arctan\frac32\in]0,\pi/2[$ (so that $\cos\alpha$ and $\sin\alpha$ are both positive). From $\cos^2\alpha+\sin^2\alpha=1$, you get $1+\tan^2\alpha=\dfrac1{\cos^2\alpha}$ then

$$\cos\alpha=\frac{1}{\sqrt{1+\tan^2\alpha}}=\frac{2}{\sqrt{13}}$$ $$\sin\alpha=\tan\alpha\cos\alpha=\frac{3}{\sqrt{13}}$$

Hence $x=-\dfrac{4}{\sqrt{13}}$ and $y=\dfrac{9}{\sqrt{13}}$.