Question

if 2x+y+k=0 is a normal to the parabola x^2=16y then the value of k is __?
PLEASE HELP. I WILL MARK AS BRAINLIEST

Answer 1

value of k = -9

2x + y + k = 0 is a normal to the parabola x² = 16y. so we have to find value of k.

first find slope of tangent of parabola, dy/dx

differentiating curve with respect to x,

2x = 16 dy/dx ⇒dy/dx = x/8

so, slope of normal to the curve = -1/slope of tangent

= -8/x ......(1)

2x + y + k is a normal to the parabola.

so, slope of normal to the parabola = -2/1 ......(2)

from equations (1) and (2),

-8/x = -2/1 ⇒x = 4

and then y = x²/16 = (4)²/16 = 1

now putting, x = 4 and y = 1 in equation 2x + y + k = 0

so, k = -(2x + y) = -(2 × 4 + 1 ) = -9

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Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

The slope of the normal at any point (h, k) to the curve

$\displaystyle \: \sf{ \: y = f(x) \: \: is \: \: \ - \bigg[ \frac{dx}{dy} \bigg]_ {(h, k )} }$

GIVEN

$2x + y + k = 0 \: \:$

is a normal to the parabola

${x}^{2} = 16y$

TO DETERMINE

The value of k

CALCULATION

Let

$2x + y + k = 0 \: \: ....(1)$

is a normal to the parabola

${x}^{2} = 16y \: .....(2) \:$

$at \: the \: point \: (x_1,y_1)$

Differentiating both sides of Equation (2) with respect to x we get

$\displaystyle \: 2x = 16 \times \frac{dy}{dx}$

$\implies \: \displaystyle \: \frac{dy}{dx} = \frac{x}{8}$

$\implies \: \displaystyle \: \frac{dx}{dy} = \frac{8}{x}$

So

$\displaystyle \: \bigg[ \frac{dx}{dy} \bigg]_ {(x_1, y_1)} = \frac{8}{x_1}$

So the slope of normal the curve at the point

$(x_1, y_1) \: \: is \: \displaystyle \: - \frac{8}{x_1}$

Again representing the equation of the line (1) is slope - intercept form we get

$y = - 2x - k \: \: .....(3)$

So the slope of the line (1) is - 2

So by the given condition

$\: \displaystyle \: - \frac{8}{x_1} = - 2$

$\implies \: \: {x_1} = 4$

Now the point

$(x_1, y_1)$

lies on the curve (2)

So

${x_1}^{2} = 16 \: y_1$

$\implies \: 16 = 16 y_1$

$\implies \: y_1 = 1$

So the point is ( 4, 1)

So the point also lies on the line (1)

So

$(2 \times 4) + 1 + k = 0$

$\implies \: 9+ k = 0$

$\implies \: k = - 9$

RESULT

So the required value is

$\boxed{ \sf{ \: k = - 9 \: } \: }$