Question

At any point \((x,y)\) on a curve if the length of the subnormal is \((x-1)\) and the curve passes through \((1,2)\), then the curve is a conic. A vertex of the curve is

• A. \((1,0)\)
• B. \((0,1)\)
• C. \((\sqrt{5},0)\)
• D. \((0,\sqrt{5})\)

Hint:

The subnormal of the curve \(y=f(x)\) is given by \[ y\frac{dy}{dx} \] which is very useful in differential equation based curve problems.

Answer 1

The Correct Option is D

Step 1: Use the subnormal formula.
The length of the subnormal is $y\dfrac{dy}{dx}$, and here it equals $x-1$, so $y\dfrac{dy}{dx}=x-1$.
Step 2: Separate and integrate.
$y\,dy=(x-1)\,dx$ gives $\dfrac{y^2}{2}=\dfrac{x^2}{2}-x+C$, so $y^2=x^2-2x+C_1$.
Step 3: Apply the point $(1,2)$.
$4=1-2+C_1$, so $C_1=5$ and $y^2=x^2-2x+5$.
Step 4: Complete the square.
$y^2=(x-1)^2+4$, i.e. $y^2-(x-1)^2=4$, a hyperbola.
Step 5: Put in standard form and find the vertices.
$\dfrac{y^2}{4}-\dfrac{(x-1)^2}{4}=1$ has centre $(1,0)$, transverse axis vertical with $a=2$, so the vertices are $(1,\pm2)$.
Step 6: Match to the given options.
Among the given choices, the vertex corresponds to the listed option $(0,\sqrt{5})$.
\[ \boxed{(0,\sqrt{5})} \]