## Problems in parabola

In this page 'Problems in parabola' we are going to see some problems about tangents, normal and chord of contact to the parabola.

Problem 1:

Find the equation of the tangent to the parabola y² = 8x, which is perpendicular to the line x-2y+6=0.

Solution:

y² = 8x

= 4(2)x

So here a = 2.

Slope of the given line x-2y+6 = 0 is

2y = x+6

y = x/2 +3

So slope of the given line is 1/2.

Since the tangent is perpendicular to the line, the slope of the tangent is   -2.

Equation of the tangent whose slope is 'm'

y = mx+ a/m

Here m= -2 and a =2

Equation of tangent is

y = -2x +2/(-2)

2x+y+1=0 is the required equation.

Problem 2:

Find the equations of normals to the parabola y² =4ax at the end of the latus rectum.

Solution;

Latus rectum is the line drawn through the focus. So we have x=a.

y²= 4a(x)

y² =  4a²

y = ± 2a

So the latus rectum meets the parabola at (a,2a) and (a,-2a).

Equation of normal at (x₁,y₁) to the parabola is

y-y₁ = -y₁/2a (x-x₁)

So here, the equation of normal at (a,2a) is

y-2a = (-2a/2a) (x-a)

x+y-3a=0 is the  equation of normal at (a,2a).

Equation of normal at (a,-2a) is

y-(-2a) = [-(-2a)/2a] (x-a)

y+2a  =  x-a

x-y-3a = 0. is the equation of normal (a,-2a).

Problem 3:

Show that the chord of contact of tangents drawn from any point on the directrix passes through the focus of the parabola.

Solution:

Let us take the equation of the parabola as y ² =4ax. So the equation of the directrix is x+a =0. The focus of the parabola is (a,0).

Any point on the directrix will be (-a,k).

We know the chord of contact equation is yy₁ = 2a(x+x₁)

At (-a,k) the chord of contact is              yk  = 2a(x-a)

The focus (a,0) satisfies this equation.

[0k = 2a(a-a) = 0].

This proves the chord of contact yk  = 2a(x-a) passes through the focus.

Problem 4:

Find the locus of the point whose chords of contact of tangents to the parabola y² =8x subtend a right angle at the vertex.

Solution:

Let the moving point be P(x₁,y₁).

Then the chord of contact is at P is yy₁ = 4(x+x₁)

Homogenizing the equation of the parabola with the above equation

y²   =  8x [(yy₁-4x)/4x₁]

4x₁y²   =   8y₁xy-32x²

32x² -8y₁xy +4x₁y² =0

This equation represents the pair of lines joining the vertex to the intersection of chord of contact and the parabola.

Angle between these lines is given to be 90⁰.

Coefficient of x² + coefficient of y² =0

32 + 4x₁ = 0

Locus of P is 32 + 4x = 0

x  = -8 is the required locus.