Let the normal at a point $P$ on the curve $y^{2}-3 x^{2}+y+10=0$ intersect the $y$ - axis at $\left(0, \frac{3}{2}\right)$. If m is the slope of the tangent at P to the curve, then $|\mathrm{m}|$ is equal to

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Application of Derivatives

Let the normal at a point

P

on the curve

y^{2} - 3 x^{2} + y + 10 = 0

intersect the

y

- axis at

(0, \frac{3}{2})

. If m is the slope of the tangent at P to the curve, then

∣ m ∣

is equal to

Answer: 4

Let

P (x_{1}, y_{1})

lie on the curve. Implicit differentiation gives

2 y y^{'} - 6 x + y^{'} = 0

, so the tangent slope is

m = \frac{6 x _{1}}{2 y _{1} + 1}

and the normal slope is

- \frac{2 y _{1} + 1}{6 x _{1}}

. Since the normal through

P

passes through

(0, \frac{3}{2})

\frac{\frac{3}{2} - y _{1}}{- x _{1}} = - \frac{2 y _{1} + 1}{6 x _{1}}

, which simplifies to

9 - 6 y_{1} = 2 y_{1} + 1

, so

y_{1} = 1

. Putting

y_{1} = 1

into the curve gives

x_{1}^{2} = 4

, hence

x_{1} = \pm 2

, and the slope is

m = \frac{6 ( \pm 2 )}{3} = \pm 4

, so

∣ m ∣ = 4

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Integer answer