The equation of the normal to the parabola y² = 16 x with slope – 1/4 is-

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.

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 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 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 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)

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.

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

An element, X has the following isotopic composition, the weighted average atomic mass of the naturally- occurring element 'X' is closest to :

The line x + my + n = 0 will touch the parabola y² = 4ax, if

a quadratic equation has equal roots if the D value is zero. this is the condition of the line touching the curve.