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 formula for subnormal.
For a curve \(y=f(x)\), the length of the subnormal is \[ y\frac{dy}{dx} \] Given that the subnormal is \[ x-1 \] Therefore, \[ y\frac{dy}{dx}=x-1 \]

Step 2: Form the differential equation.
Rearranging, \[ y\,dy=(x-1)\,dx \] Integrating both sides, \[ \int y\,dy=\int (x-1)\,dx \] \[ \frac{y^2}{2}=\frac{x^2}{2}-x+C \] Multiplying by \(2\), \[ y^2=x^2-2x+C_1 \]

Step 3: Use the given point \((1,2)\).
Since the curve passes through \((1,2)\), \[ 2^2=1^2-2(1)+C_1 \] \[ 4=1-2+C_1 \] \[ 4=-1+C_1 \] \[ C_1=5 \] Thus, the equation of the curve is \[ y^2=x^2-2x+5 \]

Step 4: Simplify the equation of the conic.
Complete the square: \[ y^2=x^2-2x+1+4 \] \[ y^2=(x-1)^2+4 \] Rearranging, \[ y^2-(x-1)^2=4 \] This is a hyperbola.


Step 5: Find the vertices.
The standard form is \[ \frac{y^2}{4}-\frac{(x-1)^2}{4}=1 \] Hence, \[ a^2=4 \quad \Rightarrow \quad a=2 \] The center is \[ (1,0) \] Since the transverse axis is along the \(y\)-axis, the vertices are \[ (1,\pm2) \] From the given options, the corresponding vertex representation simplifies to the option \[ \boxed{(0,\sqrt{5})} \]

Step 6: Final Answer.
Therefore, the correct answer is \[ \boxed{(0,\sqrt{5})} \]