Question

The length of a focal chord of the parabola $y^2=4ax$ at a distance b from the vertex is c. Then.

• A. $a^2=bc$
• B. $a^3=b^{2}c$
• C. $b^2=ac$
• D. $b^{2}c=4a^3$

Answer 1

The correct option is D $b^{2}c=4a^3$

Parabola $P:y²=4ax--\left(1\right)$

Vertex $=O(0,0)$ Focus: $F(a,0)$

Let the Focal chord $L$ be $(y-0)=m(x-a)$

So $y=mx-ma--\left(2\right)$\

Given b = Distance of O from L.

$=>b²=m²a²/(1+m²)---\left(3\right)$

Find intersections $A(x_1,y_1),B(x_2,y_2)$ of P & L:

Eliminate y from (1) & (2):

$m²x²-2amx+m²a²=4ax$

$m²x²-2a(m+2)x+m²a²=0---\left(3\right)$

$x_1,x_2$ are the roots, $x_1+x_2=2a(m+2)/m²and{x}_1{x}_2=a²--\left(4\right)$

Eliminate x from (1) & (2):

$y=m\left(y²/4a\right)-ma$

$my²-4ay-4a²m=0--\left(5\right)$

$y_1,y_2$ are the roots, $y_1+y_2=4a/mand{y}_1{y}_2=-4a²---\left(6\right)$

Now Length of focal chord $=c=AB$

$c²=(x_1-x_2)²+(y_1-y_2)²$

$=(x_1+x_2)²-4x_1{x}_2+(y_1+y_2)²-4y_1{y}_2$

$=4a²(m⁴+4+4m²)/m⁴-4a²+16a²/m²+16a²$

$=16a²(m²+1)²/m⁴$

$=16a²\left(a²/b²\right)²---using\left(3\right)$

$=>b²c=4|a³|$ , Modulus as $"a"$can be $+veor-ve.$

Hence, the option $D$ is the correct answer.