Question

Let A be the vertex and L be the length of the latus rectum of the parabola, $y^2-2y-4x-7=0.$ The equation of the parabola with a as vertex 2 L the length of the latus rectum and axis at right angles that of the given curve is

• A. $x^2+4x+8y-4=0$
• B. $x^2+4x+8y-12=0$
• C. $x^2+4x-8y+12=0$
• D. $x^2+8x-4y+8=0$

Answer 1

The correct option is C $x^2+4x-8y+12=0$

$P_1:y^2-2y-4x-7=0....\left(i\right)$

$=>{(y-1)}^2=4x+8$

Vertex: $A:(-2,1)$

Now axis of $\left(i\right)$, $y-1=\mathrm{0..........}\left(ii\right)$

for new Parabola $P_2:Axis=>x=\lambda .....\left(iii\right)$

as $\left(iii\right)$ passes through $A=>x=-\mathrm{2.....}\left(iv\right)$

$L.R.of{L}_2=2L_1=8$

$=>a_2=2$

and vertex $=>(-2,1)$

$=>$ equation of $p_2:(x-2)=-4\left(2\right)(y-1)$

$=>x^2+4+4x=-(8y-8)$

$=>{4x}^2+4x+8y-4=\mathrm{0........}\left(v\right)$

and also,

$=>{(x+2)}^2=4\left(2\right)(y-1)$

$=>x^2+4+4x-8y+8=0$

$=>x^2+4x-8y+12=\mathrm{0......}\left(vi\right)$

So, Equation $\left(v\right)$ is in the option.