JEE Main 2020 (Online) 8th January Morning ShiftQuestion 1
Let the normal at a point $P$ on the curve $y^2-3x^2+y+10=0$ intersect the $y$-axis at $\left(0,\frac32\right)$. If $m$ is the slope of the tangent at $P$ to the curve, then $|m|$ is equal to ________.
Differentiation and its Application
Answer$4$
Solution
Concept. The normal at $P(x_0,y_0)$ passes through $\left(0,\frac32\right)$; this gives $y_0$, and the curve gives $x_0$. Formulas used. (1) $2yy'-6x+y'=0$ $\Rightarrow$ $m=\dfrac{6x}{2y+1}$Step 1 - Condition. The slope of the line from $P$ to $\left(0,\frac32\right)$ is $-\frac1m$: $$\frac{y_0-\frac32}{x_0}=-\frac{2y_0+1}{6x_0}\ \Longrightarrow\ 6y_0-9=-2y_0-1\ \Longrightarrow\ y_0=1.$$ Step 2 - $x_0$. On the curve: $1-3x_0^2+1+10=0$, so $x_0^2=4$. Step 3 - Slope. By (1), $|m|=\frac{6\cdot2}{3}=4$. Answer: 4