UPSC 2023 Maths Optional Paper 1 Q4c — Step-by-Step Solution

15 marks · Section A

Straight lines in 3D · Analytic Geometry · asked 5× in 13 yrs · Read the full method →

Question

Prove that the locus of a line which meets the lines $y=mx,\,z=c$ ; $y=-mx,\,z=-c$ and the circle $x^2+y^2=a^2,\,z=0$ is

c^2 m^2(cy-mzx)^2+c^2(yz-cmx)^2=a^2 m^2(z^2-c^2)^2.

Technique

Parametrise the variable line by its $z$ -coordinate; use that the $z=0$ point is the midpoint of $A,B$ on the two given skew lines; the circle constraint $(\star)$ then translates into the desired locus equation.

Solution

Step 1 — Parametrise points on each given line/circle.

$L_1:\,y=mx,\,z=c\Rightarrow$ a point $A=(\alpha,m\alpha,c)$ for parameter $\alpha$ .

$L_2:\,y=-mx,\,z=-c\Rightarrow$ a point $B=(\beta,-m\beta,-c)$ for parameter $\beta$ .

The midpoint of $AB$ at $z=0$ : indeed, $\bigl((\alpha+\beta)/2,(m\alpha-m\beta)/2,0\bigr)$ . This $z=0$ midpoint must coincide with a point of the circle $C=(a\cos\phi,a\sin\phi,0)$ on the line through $A,B$ — but only if the line passes through the circle. We use this midpoint condition.

Step 2 — Use the fact that the variable line meets all three.

A line that meets $L_1$ at $A$ and $L_2$ at $B$ has its third intersection with the plane $z=0$ at the midpoint of $AB$ (because $A$ has $z=c$ , $B$ has $z=-c$ , and the line is straight in $z$ , so its $z=0$ point is the midpoint). For this midpoint to lie on the circle $x^2+y^2=a^2$ :

\Bigl(\frac{\alpha+\beta}{2}\Bigr)^2+\Bigl(\frac{m(\alpha-\beta)}{2}\Bigr)^2=a^2,

i.e.,

(\alpha+\beta)^2+m^2(\alpha-\beta)^2=4a^2.\qquad(\star)

Step 3 — Parameterise points on the variable line.

Take a generic point $P=(x,y,z)$ on the line. Use the parameter $s=(z\text{-coord scaled})$ . With $z$ ranging linearly from $c$ at $A$ to $-c$ at $B$ , set $s=z/c\in[-1,1]$ (with $s=1$ at $A,\,s=-1$ at $B,\,s=0$ at the midpoint). The line: $(1+s)/2$ part $A$ + $(1-s)/2$ part $B$ — so

x=\frac{(1+s)\alpha+(1-s)\beta}{2}=\frac{(\alpha+\beta)+s(\alpha-\beta)}{2}.

y=\frac{(1+s)m\alpha-(1-s)m\beta}{2}=\frac{m[(\alpha-\beta)+s(\alpha+\beta)]}{2}.

Let $p=\alpha+\beta,\,q=\alpha-\beta$ . Then

x=\tfrac{p+sq}{2},\quad y=\tfrac{m(q+sp)}{2},\quad s=z/c.

Step 4 — Solve for $p,q$ .

From $x$ : $p=2x-sq$ . Substitute into $y$ : $y=\tfrac{m}{2}(q+s(2x-sq))=\tfrac{m}{2}(q+2sx-s^2q)=\tfrac{m}{2}(q(1-s^2)+2sx)$ .

Solve for $q$ : $\dfrac{2y}{m}=q(1-s^2)+2sx\Rightarrow q=\dfrac{2(y/m-sx)}{1-s^2}=\dfrac{2(y-msx)}{m(1-s^2)}$ .

With $s=z/c$ , $1-s^2=(c^2-z^2)/c^2$ , $sm=mz/c$ :

q=\frac{2(y-mzx/c)}{m(c^2-z^2)/c^2}=\frac{2c(cy-mzx)}{m(c^2-z^2)}.

And $p=2x-sq=2x-\dfrac{z}{c}\cdot\dfrac{2c(cy-mzx)}{m(c^2-z^2)}=2x-\dfrac{2z(cy-mzx)}{m(c^2-z^2)}$ .

Combine: $p=\dfrac{2xm(c^2-z^2)-2z(cy-mzx)}{m(c^2-z^2)}=\dfrac{2xmc^2-2xmz^2-2zcy+2mz^2x}{m(c^2-z^2)}=\dfrac{2c(cmx-yz)}{m(c^2-z^2)}$ .

Step 5 — Apply the circle constraint $(\star)$ : $p^2+m^2q^2=4a^2$ .

p^2=\frac{4c^2(cmx-yz)^2}{m^2(c^2-z^2)^2},\quad m^2 q^2=m^2\cdot\frac{4c^2(cy-mzx)^2}{m^2(c^2-z^2)^2}=\frac{4c^2(cy-mzx)^2}{(c^2-z^2)^2}.

Sum equals $4a^2$ :

\frac{4c^2(cmx-yz)^2}{m^2(c^2-z^2)^2}+\frac{4c^2(cy-mzx)^2}{(c^2-z^2)^2}=4a^2.

Multiply through by $m^2(c^2-z^2)^2$ :

c^2(cmx-yz)^2+c^2 m^2(cy-mzx)^2=a^2 m^2(c^2-z^2)^2.

Note $(cmx-yz)^2=(yz-cmx)^2$ and $(c^2-z^2)^2=(z^2-c^2)^2$ :

Answer

\boxed{\;c^2 m^2(cy-mzx)^2+c^2(yz-cmx)^2=a^2 m^2(z^2-c^2)^2.\;}