Question

Find the equations of tangents and normal at the extremities of latus rectum of the parabola $y^2=4ax.$

• A. Equation of tangent is y = x + a
• B. Equation of tangent is y = -x - a
• C. Equation of normal is y = -x + 3a
• D. Equation of normal is y = x - 3a

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A

Equation of tangent is y = x + a

B

Equation of tangent is y = -x - a

C

Equation of normal is y = -x + 3a

D

Equation of normal is y = x - 3a

Let the parabola be $y^2=4ax$ and extremities of latus

rectum are (a,2a) $\&$ (a,-2a)

P is the point of intersection of two tangents

Equation of tangent at A

$y{y}_1=2a(x+x_1)$

$y\left(2a\right)=2a(x+a)$

$y=x+a$ .........(1)

Equation of tangent at B

$y(-2a)=2a(x+a)$

$y=-x-a$ .........(2)

Now, we shall calculate the equation of normal at point A

Slope of normal $=-\frac{1}{slopeoftangent}=-\frac{1}{1}=-1$

Equation of normal at point A

$y-2a=-1(x-a)$

$y=-x+a+2a$

$y=-x+3a$ .........(3)

Slope of normal at through point B

$=\frac{-1}{-1}=1$

Equation of normal at point B

$y-(-2a)=1(x-a)$

$y=x-a-2a$

$y=x-3a$ .........(4)