Advertisements
Advertisements
Question
Find the equation of tangents to the curve y= x3 + 2x – 4, which are perpendicular to line x + 14y + 3 = 0.
Solution
Consider the given equation,
y = x3+2x-4
Differentiating the above function with respect to x, we have,
`dy/dx=3x^2+2`
⇒ m1 = 3x2+2
Given that the tangents to the given curve are perpendicular to the line x+14y+3=0
Slope of this line, `m_2=(-1)/14`
Since the given line and the tangents to the given curve are perpendicular, we have,
m1 x m2 = -1
`=>(3x^2+2)((-1)/14)=-1`
⇒ 3x2 + 2 = 14
⇒ 3x2 = 12
⇒ x2 = 4
⇒ x = ±2
if x =2, y=x3 + 2x -4
⇒ y = 23 + 2 x 2 - 4
⇒ y = 8
if x = -2, y =x3 + 2x -4
⇒ y = (-2)3+ 2 x (-2) - 4
⇒ y = -16
Equation of the tangent having slope m at the point (x1,y1) is (y-y1)=m(x-x1)
Equation of the tangent at P(2,8) with slope 14
(y-8)=14(x-2)
⇒ y -8 = 14x -28
⇒ 14x -y= 20
Equation of the tangent at P(-2,-16) with slope 14
(y+1=6) = 14(x+2)
⇒ y +16 = 14x +28
⇒ 14x - y = -12
APPEARS IN
RELATED QUESTIONS
Find the slope of the tangent to the curve y = x3 − 3x + 2 at the point whose x-coordinate is 3.
Find points on the curve `x^2/9 + "y"^2/16 = 1` at which the tangent is parallel to x-axis.
Find the equations of the tangent and normal to the given curves at the indicated points:
y = x4 − 6x3 + 13x2 − 10x + 5 at (0, 5)
Find the points on the curve x2 + y2 − 2x − 3 = 0 at which the tangents are parallel to the x-axis.
Find the points on the curve y = `4x^3 - 3x + 5` at which the equation of the tangent is parallel to the x-axis.
Find the slope of the tangent and the normal to the following curve at the indicted point x = a cos3 θ, y = a sin3 θ at θ = π/4 ?
Find the equation of the tangent and the normal to the following curve at the indicated point y = x4 − 6x3 + 13x2 − 10x + 5 at x = 1?
Find the equation of the tangent and the normal to the following curve at the indicated point y = 2x2 − 3x − 1 at (1, −2) ?
Find the equation of the tangent and the normal to the following curve at the indicated point y2 = 4ax at \[\left( \frac{a}{m^2}, \frac{2a}{m} \right)\] ?
Find the equation of the tangent and the normal to the following curve at the indicated point y2 = 4ax at (x1, y1)?
Find the equation of the tangent and the normal to the following curve at the indicated points \[x = \frac{2 a t^2}{1 + t^2}, y = \frac{2 a t^3}{1 + t^2}\text { at } t = \frac{1}{2}\] ?
Find the equation of the tangent and the normal to the following curve at the indicated points x = asect, y = btant at t ?
Find the equations of all lines of slope zero and that are tangent to the curve \[y = \frac{1}{x^2 - 2x + 3}\] ?
Find the angle of intersection of the following curve y = 4 − x2 and y = x2 ?
Show that the following curve intersect orthogonally at the indicated point y2 = 8x and 2x2 + y2 = 10 at \[\left( 1, 2\sqrt{2} \right)\] ?
Prove that the curves y2 = 4x and x2 + y2 - 6x + 1 = 0 touch each other at the point (1, 2) ?
Find the point on the curve y = x2 − 2x + 3, where the tangent is parallel to x-axis ?
If the tangent line at a point (x, y) on the curve y = f(x) is parallel to y-axis, find the value of \[\frac{dx}{dy}\] ?
Find the slope of the normal at the point 't' on the curve \[x = \frac{1}{t}, y = t\] ?
Find the coordinates of the point on the curve y2 = 3 − 4x where tangent is parallel to the line 2x + y− 2 = 0 ?
If the tangent to the curve x = a t2, y = 2 at is perpendicular to x-axis, then its point of contact is _____________ .
The slope of the tangent to the curve x = t2 + 3 t − 8, y = 2t2 − 2t − 5 at point (2, −1) is ________________ .
Show that the equation of normal at any point on the curve x = 3cos θ – cos3θ, y = 3sinθ – sin3θ is 4 (y cos3θ – x sin3θ) = 3 sin 4θ
Find the condition that the curves 2x = y2 and 2xy = k intersect orthogonally.
Find the angle of intersection of the curves y = 4 – x2 and y = x2.
The slope of tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2, –1) is ______.
The two curves x3 – 3xy2 + 2 = 0 and 3x2y – y3 – 2 = 0 intersect at an angle of ______.
`"sin"^"p" theta "cos"^"q" theta` attains a maximum, when `theta` = ____________.
Tangents to the curve x2 + y2 = 2 at the points (1, 1) and (-1, 1) are ____________.
The slope of the tangentto the curve `x= t^2 + 3t - 8, y = 2t^2 - 2t - 5` at the point `(2, -1)` is
