Question

Intersection points of tangents and normal at the extremities of latus rectum of the parabola $y^2=4ax$ will form a square.

• A. True
• B. False

Answer 1

The correct option is A True

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)

For intersection point of tangent,

Solving equation (1) $\&$ (2)

x+a=x-a

2x=-2a

x=-a,y=0

Coordinates of Point P(-a,0)

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)

Soving equation (3) $\&$ (4)

We get, x=-3a

y=0

Coordinatenof point L (-3a,0)

Coordinates of point A (a,2a), P(-a,0), B(a,-2a) L(3a,0)

we need to check whether APBL is square

So, calculating the length of $AP=\sqrt{(a+a{)}^2+(2a{)}^2}=2\sqrt{2}a$

$PB=\sqrt{(-a-a{)}^2+(-2a{)}^2}=2\sqrt{2}a$

$BL=\sqrt{(a-3a{)}^2+(-2a{)}^2}=2\sqrt{2}a$

$AL=\sqrt{(3a-a{)}^2+(2a{)}^2}=2\sqrt{2}a$

In square diagonals of the square bisects each other.

Here, we see. Intersection points of diagonals

$M(\frac{-a+3}{2},0)=M(\frac{a+a}{2},\frac{2a-2a}{2})=M(a,0)$

and

$PM=ML=AM=BM$

So,APBL is a square.