Question

The number of normals drawn to the parabola \(y^2=4x\) from the point \((1,0)\) is

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(3\)

Hint:

To find number of normals from a point to a parabola, write normal in parametric form and solve for real values of parameter \(t\).

Answer 1

The Correct Option is B

Step 1: Parametric point on parabola.
For \(y^2=4x\), parametric form is:
\[ (x,y)=(t^2,2t) \]
Step 2: Slope of tangent and normal.
Differentiate:
\[ 2y\frac{dy}{dx}=4 \Rightarrow \frac{dy}{dx}=\frac{2}{y} \]
At point \((t^2,2t)\):
\[ \frac{dy}{dx}=\frac{2}{2t}=\frac{1}{t} \]
So slope of normal:
\[ m_n=-t \]
Step 3: Equation of normal at parameter \(t\).
Normal passing through \((t^2,2t)\):
\[ y-2t=-t(x-t^2) \]
Step 4: Impose condition that it passes through \((1,0)\).
Put \(x=1,y=0\):
\[ 0-2t=-t(1-t^2) \]
\[ -2t=-t+t^3 \]
\[ t^3+t=0 \Rightarrow t(t^2+1)=0 \]
Step 5: Count real solutions.
\[ t=0 \;\text{(real)} \]
\[ t^2+1=0 \Rightarrow t=\pm i\;\text{(imaginary)} \]
Only one real value of \(t\).
Final Answer:
\[ \boxed{1} \]