Question
The equations of the normal at the ends of the latus rectum of the parabola y2 = 4ax are given by-
- x2 – y2 – 6ax + 9a2 = 0
- x2 – y2 – 6ay + 9a2 = 0
- x2 – y2 – 6ax – 6ay + 9a2 = 0
- None of these
Hint:
find the equations of normal at the ends of latus rectum and multiply the two to get the general equation.
The correct answer is: x2 – y2 – 6ax + 9a2 = 0
x2-y2-6ax+9a2=0
the end points of the latus rectum of the parabola are : (a,2a)and (a,-2a).
slope of tangent : dy/dx = 2/ay
slope of normal = -ay/2
equation of normal at the said points:
at (a,2a): y-2a= (-a2)(x-a)
x+y-3a=0
at (a,-2a): x-y-3a=0
general equation of both these equations:
x2-y2-6ax+9a2=0
the slope of normal = -1/ slope of tangent.
Equation of both normals can be generalized by multiplying both the equations.
Related Questions to study
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-
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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
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equation of normal of a parabola is given by:
y = -tx + 2at + at3
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Area of triangle formed by the tangents at three points t1, t2 and t3 of the parabola y2 = 4ax is-
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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
xx1=(y+y1)/2
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 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
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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-
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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
At which point the line x = my + 
touches the parabola x2 =4ay
the point of contact is the point of intersection between the 2 curves, the parabola and the line. this can be obtained by solving the two equations.
At which point the line x = my + touches the parabola x2 =4ay
the point of contact is the point of intersection between the 2 curves, the parabola and the line. this can be obtained by solving the two equations.
For what value of k, the line 2y – x + k = 0 touches the parabola x2 + 4y = 0-
when a quadratic equation has 2 equal roots, the D value is 0 . D= b^2 - 4ac.=0
this gives us the possible value(s) of k
For what value of k, the line 2y – x + k = 0 touches the parabola x2 + 4y = 0-
when a quadratic equation has 2 equal roots, the D value is 0 . D= b^2 - 4ac.=0
this gives us the possible value(s) of k
