Question

If the tangent & normal at any point 'P' of the parabola $y^2=4ax$ intersect the axis of the parabola at $T\&G$ then $ST=SG\ne SP.$

• A. True
• B. False

Answer 1

The correct option is B

False

Let the point P be $(a{t}^2,2at)$. The equation of tangent at point P

$T=0$

$y.\left(2at\right)-2a(x+a{t}^2)=0$

$ty-x-a{t}^2=0$ .......(1)

We need to find the equation of normal which passes through point $P(a{t}^2,2at)$

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

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

Equation of Normal is $y-2at=-t(x-a{t}^2)$

$y=-tx+2at+a{t}^3$ ........(2)

For coordinates of point T:

since, T lies on the x-axis y=0

Substituting y=0 in equation of tangent

$t\times 0-x-a{t}^2=0$

$x=-a{t}^2$

$T(-a{t}^2,0)$

For coordinates of point G

G also lies on the x-axis,y=0

Substituting y=0 in the equation of normal

$0=-tx+2at+a{t}^3$

$x=\frac{2at+a{t}^3}{t}=2a+a{t}^2$

$G(2a+a{t}^2,0)$

Coordinates of S focus=(a,0)

Now, we need to check for SP,ST and SG

So, $SP=\sqrt{(a{t}^2-a{)}^2+(2at-a{)}^2}$

$=\sqrt{a^2{t}^2+a^2-2a^2{t}^2+4a^2{t}^2}$

$=\sqrt{a^2{t}^2+a^2+2a^2{t}^2}$

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

$SP=|a{t}^2+a|$

$ST=\sqrt{(-a{t}^2-a{)}^2+0}$

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

$ST=|a{t}^2+a|$

$SG=\sqrt{(2a+a{t}^2-a{)}^2+0}$

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

$=|a{t}^2+a|$

So, we get,

$SP=ST=SG=|a{t}^2+a|$

So, the given statement is not correct.