Question

If the normal to the curve$y^2=5x–1$, at the point $(1,–2)$ is of the form$ax–5y+b=0$, then $a$ and $b$ are

• A. $4,–14$
• B. $4,14$
• C. $–4,14$
• D. $–4,–14$

Answer 1

The correct option is A

$4,–14$

The explanation for the correct answer.

Step 1. Find the slope of the normal

Given: $y^2=5x–1$

Differentiate given with respect to $x$

$$\begin{gathered} \Rightarrow 2y\frac{dy}{dx}=5 \\ \Rightarrow \frac{dy}{dx}=\frac{5}{2y} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(1,-2\right)}=\frac{5}{-4} \end{gathered}$$

The slope of the normal is $\frac{-1}{{\displaystyle \raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dx$}\right.}}=\frac{-1}{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$-4$}\right.}}=\frac{4}{5}$

Step 2. Find the equation of the normal

The equation of normal is given as $y-y_1=m\left(x-x_1\right)$

$$\begin{gathered} \Rightarrow y-(-2)=\frac{4}{5}\left(x-1\right) \\ \Rightarrow 5(y+2)=4(x-1) \\ \Rightarrow 5y+10=4x-4 \\ \Rightarrow 4x-5y-14=0 \end{gathered}$$

Compare the above equation with $ax–5y+b=0$

$a=4,b=-14$

Hence option (a) is correct