What is the slope of the normal at the point (at2, 2at) of the parabola y2 = 4ax ?

Question

What is the slope of the normal at the point (at2, 2at) of the parabola y2 = 4ax ? (A) (1/t) (B) t (C) - t (D) - (1/t)

Tardigrade Admin · Accepted Answer

Equation of parabola is y2 = 4ax ⇒ 2y (dy/dx) = 4a (On differentiating w.r.t âxâ ) ∴ (dy/dx) = (2a/y), [slope of tangent] So, slope of normal = - ((dx/dy) )at2 , 2 at = - ((y/2a)) = - (2at/2a) = - t

Q. What is the slope of the normal at the point $(a t^{2}, 2 a t)$ of the parabola $y^{2} = 4 a x$ ?

$\frac{1}{t}$

t

- t

$- \frac{1}{t}$

Equation of parabola is $y^{2} = 4 a x$
$\Rightarrow 2 y \frac{d y}{d x} = 4 a$ (On differentiating w.r.t ‘x’ )
$∴ \frac{d y}{d x} = \frac{2 a}{y},$ [slope of tangent]
So, slope of normal $= - (\frac{d x}{d y})_{a t^{2}, 2 a t}$
$= - (\frac{y}{2 a}) = - \frac{2 a t}{2 a} = - t$

Q. What is the slope of the normal at the point (at2,2at) of the parabola y2=4ax ?

t1​

t

- t

−t1​

Equation of parabola is y2=4ax ⇒2ydxdy​=4a (On differentiating w.r.t ‘x’ ) ∴dxdy​=y2a​, [slope of tangent] So, slope of normal =−(dydx​)at2,2at​ =−(2ay​)=−2a2at​=−t

Q. What is the slope of the normal at the point $(a t^{2}, 2 a t)$ of the parabola $y^{2} = 4 a x$ ?

$\frac{1}{t}$

$- \frac{1}{t}$

Equation of parabola is $y^{2} = 4 a x$
$\Rightarrow 2 y \frac{d y}{d x} = 4 a$ (On differentiating w.r.t ‘x’ )
$∴ \frac{d y}{d x} = \frac{2 a}{y},$ [slope of tangent]
So, slope of normal $= - (\frac{d x}{d y})_{a t^{2}, 2 a t}$
$= - (\frac{y}{2 a}) = - \frac{2 a t}{2 a} = - t$