Question

Number of distinct normals that can be drawn from $(\frac{11}{4},\frac{1}{4})$ to the parabola $y^2=4x$ is

• A. $3$
• B. $2$
• C. $1$
• D. $4$

Answer 1

The correct option is C $1$

$\begin{array}{c}{y}^2=4x\\ d=1\\ N:y=mx-2am-a{m}^3\\ y=mx-2m-m^3\\ at\left(\frac{11}{4},\frac{1}{4}\right)\\ \Rightarrow \frac{1}{4}=\frac{11m}{4}-2m-3m^2\\ \Rightarrow \frac{1}{4}=\frac{11m-8m-4m^3}{4}\\ \Rightarrow 1=3m-4m^3\\ \Rightarrow 4m^3-3m+1=0\\ m=-1\Rightarrow -4+3+1=0\\ \therefore \left(m+1\right)\left(4m^2-4m+1\right)=0\\ Here,D={\left(-4\right)}^2-4\times 4\times 1<0\\ So,novaluesofm\\ m+1=0\\ \therefore 4m^2-4m+1=0\\ So,m=-1\\ Hence,optionCisthecorrectanswer.\end{array}$