Question

Find the equation of the tangent to y²=16x, which is parallel to the line 4x-y=1

Answer 1

we have to find equation of the tangent to y² = 16x ,which is parallel to the line 4x - y = 1.

differentiating y² = 16x with respect to x,

i.e., 2y dy/dx = 16

⇒dy/dx = 8/y

hence, slope of tangent to the curve , y² = 16x is 8/y

now, tangent to the curve , y² = 16x is parallel to the line 4x - y = 1

so, slope of tangent = slope of line

⇒8/y = -(4)/(-1) = 4

⇒y = 2

so, slope of tangent , m = dy/dx = 8/y = 8/2 = 4

putting, y = 2 in equation y² = 16x

i.e., (2)² = 16x ⇒x = 1/4

hence, point of contact of tangent to the curve y² = 16x is (1/4, 2)

so, equation of tangent is ...

(y - y₁) = m(x - x₁)

⇒(y - 2) = 4(x - 1/4)

⇒y - 2 = 4x - 1

⇒y - 4x -1 = 0

Answer 2

Answer:

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