All ellipse normals through a fixed point · source reference

Printed question

MI Prelim 9758/2023/02/Q4. The curve $C$ has equation $2x^2+y^2=20$.

(i) Sketch curve $C$, giving the exact coordinates of any points where the curve meets the axes. [2]

(ii) Show that $\frac{dy}{dx}=-\frac{2x}{y}$. [1]

(iii) It is given that point $P(a,b)$ is on $C$. The normal to $C$ at $P$ passes through the point $(1,0)$. Find the possible coordinates of $P$. [5]

[2023/MI/Prelim/II/4]

Printed working

(i) $2x^2+y^2=20\Rightarrow\frac{x^2}{10}+\frac{y^2}{20}=1\Rightarrow\frac{x^2}{(\sqrt{10})^2}+\frac{y^2}{(\sqrt{20})^2}=1\Rightarrow\frac{x^2}{(\sqrt{10})^2}+\frac{y^2}{(2\sqrt5)^2}=1$.

The ellipse has intercepts $(\pm\sqrt{10},0)$ and $(0,\pm2\sqrt5)$.

(ii) $4x+2y\frac{dy}{dx}=0\Rightarrow2y\frac{dy}{dx}=-4x\Rightarrow\frac{dy}{dx}=-\frac{4x}{2y}=-\frac{2x}{y}$ (shown).

(iii) Method 1. Normal gradient $=y/(2x)$, at $(a,b)$ it is $b/(2a)$. Since the normal passes through $(1,0)$,

$$\frac b{2a}=\frac{b-0}{a-1}\Rightarrow ab-b=2ab\Rightarrow ab+b=0\Rightarrow b(a+1)=0.$$

So $b=0$ or $a=-1$. If $b=0$, $2a^2+0^2=20$, $a=\pm\sqrt{10}$. If $a=-1$, $2(-1)^2+b^2=20$, $b^2=18$, $b=\pm\sqrt{18}=\pm3\sqrt2$.

Points: $(\sqrt{10},0)$, $(-\sqrt{10},0)$, $(-1,-3\sqrt2)$, $(-1,3\sqrt2)$.

Method 2. Normal equation $y-b=\frac b{2a}(x-a)$ ⇒ $y=\frac b{2a}x-\frac b2+b=\frac b{2a}x+\frac b2$.

Substitute $(1,0)$: $0=\frac b{2a}(1)+\frac b2\Rightarrow0=b+ab\Rightarrow b(1+a)=0$. Again $b=0$ or $a=-1$, and the same substitutions yield the same four points.

For completeness, the excluded case $a=0$ has vertical normal $x=0$, which cannot pass through $(1,0)$; the horizontal normals at $b=0$ do pass through it.

Teacher notes

Since the coefficients of $x^2$ and $y^2$ have the same sign and different magnitudes, $C$ is an ellipse. Put the equation of $C$ in standard form.

Answer key

(i) Same labelled ellipse. (iii) $(\sqrt{10},0)$, $(-\sqrt{10},0)$, $(-1,-3\sqrt2)$ and $(-1,3\sqrt2)$.

Native diagrams