Question
The equation of the normal having slope m of the parabola y2 = x + a is
- y = mx – am – am3
- y = mx – 2am – am3
- 4y = 4mx + 4am – 2m – m3
- 4y = 4mx + 2am –am3
Hint:
equation of normal:
y= mX-2am-am^3
The correct answer is: 4y = 4mx + 4am – 2m – m3
4y = 4mx + 4am – 2m – m3
replace x+a by X
the equation becomes
y^2 = 4(1/4)(X)
therfore, equation of normal:
y= mX-2am-am^3
y= m(x+a) -2(1/4)m- (1/4)m^3
on solving, we get
4y = 4mx + 4am – 2m – m3
parametric form of slope of parabola = y+tx = 2at + at^3
Related Questions to study
Find the equation of normal to the curve 2y = 3–x2 at the point (1, 1)
equation of straight line :
(y-y1)=m(x-x1)
Find the equation of normal to the curve 2y = 3–x2 at the point (1, 1)
equation of straight line :
(y-y1)=m(x-x1)
The equations of the normal at the ends of the latus rectum of the parabola y2 = 4ax are given by-
the slope of normal = -1/ slope of tangent.
Equation of both normals can be generalized by multiplying both the equations.
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the slope of normal = -1/ slope of tangent.
Equation of both normals can be generalized by multiplying both the equations.
The line 
x + my + n = 0 is a normal to the parabola y2 = 4 ax if-
equation of normal of a parabola is given by:
y = -tx + 2at + at3
The line x + my + n = 0 is a normal to the parabola y2 = 4 ax if-
equation of normal of a parabola is given by:
y = -tx + 2at + at3
The equation of the normal to the parabola y2 = 16 x with slope – 1/4 is-
equation of normal of a parabola is given by:
y = -tx + 2at + at3
equation of normal in slope form: y = mx – 2am – am3
The equation of the normal to the parabola y2 = 16 x with slope – 1/4 is-
equation of normal of a parabola is given by:
y = -tx + 2at + at3
equation of normal in slope form: y = mx – 2am – am3
The equation of normal on point (2,4) to the parabola y2 = 8x is-
we can also use equation of normal in slope form: y = mx – 2am – am3
The equation of normal on point (2,4) to the parabola y2 = 8x is-
we can also use equation of normal in slope form: y = mx – 2am – am3
Area of triangle formed by the tangents at three points t1, t2 and t3 of the parabola y2 = 4ax is-
area under a triangle is given by the determinant of the vertex points of the triangle.
Area of triangle formed by the tangents at three points t1, t2 and t3 of the parabola y2 = 4ax is-
area under a triangle is given by the determinant of the vertex points of the triangle.
The equation of the common tangents to the parabolas y2 = 4x and x2 = 32y is-
the equation of tangent to the parabola y2= 4ax is y = mx + a/m
The equation of the common tangents to the parabolas y2 = 4x and x2 = 32y is-
the equation of tangent to the parabola y2= 4ax is y = mx + a/m
The equation of tangent to the parabola x2 = y at one extremity of latus rectum in the first quadrant is
equation of tangent : T=0 of
X2= 4ay is given by
xx1=(y+y1)/2
The equation of tangent to the parabola x2 = y at one extremity of latus rectum in the first quadrant is
equation of tangent : T=0 of
X2= 4ay is given by
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The point of contact of the line 2x – y + 2 = 0 with the parabola y2 = 16 x is-
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The point of contact of the line 2x – y + 2 = 0 with the parabola y2 = 16 x is-
for a line to be a tangent of a parabola, the value of a/m should be equal to the value of c.
The point on the curve y2 = x the tangent at which makes an angle of 45º with x-axis will be given by
the point of contact of a tangent with the curve is given by (x,y) =(a/m2, 2a/m). this is obtained by solving the equation of tangent with the equation of curve.
The point on the curve y2 = x the tangent at which makes an angle of 45º with x-axis will be given by
the point of contact of a tangent with the curve is given by (x,y) =(a/m2, 2a/m). this is obtained by solving the equation of tangent with the equation of curve.
The equation of the tangent to the parabola y = 2 + 4x –4x2 with slope –4 is
equation of a straight line : y-y1=m(x-x1)
m is given, calculate x1 and y1 by differentiating the equation of parabola and equating it with the slope.
The equation of the tangent to the parabola y = 2 + 4x –4x2 with slope –4 is
equation of a straight line : y-y1=m(x-x1)
m is given, calculate x1 and y1 by differentiating the equation of parabola and equating it with the slope.
The equation of the tangent at vertex to the parabola 4y2 + 6x = 8y + 7 is
if the slope of a line is infinite, then it is parallel to the y axis and the equation becomes x=a, where x is the x intercept.
Slope of the tangent : dy/dx
The equation of the tangent at vertex to the parabola 4y2 + 6x = 8y + 7 is
if the slope of a line is infinite, then it is parallel to the y axis and the equation becomes x=a, where x is the x intercept.
Slope of the tangent : dy/dx
The point where the line x + y = 1 touches the parabola y = x– x2 , is-
the point of intersection of two curves can be found out by solving the equations of the two curves,
The point where the line x + y = 1 touches the parabola y = x– x2 , is-
the point of intersection of two curves can be found out by solving the equations of the two curves,
The equation of the tangent to the parabola y2 = 4x at the point (1, 2) is-
Slope of tangent = dy/dx
Equation of line: y-y1 = m(x-x1)
The equation of the tangent to the parabola y2 = 4x at the point (1, 2) is-
Slope of tangent = dy/dx
Equation of line: y-y1 = m(x-x1)
If a tangent to the parabola 4y2 = x makes an angle of 60º with the x- axis, then its point of contact is-
point of contact = (a/m2, 2a/m)
slope = tan ( angle made with the x axis)
this gives us the value of m
y^2 = x/4 = 4(1/16)x
a = 1/16
If a tangent to the parabola 4y2 = x makes an angle of 60º with the x- axis, then its point of contact is-
point of contact = (a/m2, 2a/m)
slope = tan ( angle made with the x axis)
this gives us the value of m
y^2 = x/4 = 4(1/16)x
a = 1/16
