Question:

If $2x + 3y + 12 = 0$ and $x - y + 4 \lambda = 0$ are conjugate with respect to the parabola $y^2 = 8x$, then $\lambda$ is equal to

Updated On: Jun 18, 2022
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The Correct Option is D

Solution and Explanation

If $2 x +3 y +12=0$ and $x - y +4 \lambda=0$
are conjugate with respect to parabola $y ^{2}=8 x$ then $\lambda$ is equal to :
$1_{1} x +m_{1} y+ n_{1}=0$
$l _{2} x + m _{2} y + n _{2}=0$
Are conjugate with respect to parabola $y ^{2}=4 ax$
$l _{ l } n _{2}+ l _{2} n _{1}=2 am _{1} m _{2}$
$8 \lambda+12=-12$
$\lambda=-3$
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Concepts Used:

Parabola

Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).

Parabola


 

 

 

 

 

 

 

 

 

Standard Equation of a Parabola

For horizontal parabola

  • Let us consider
  • Origin (0,0) as the parabola's vertex A,
  1. Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
  2. P(x,y) as the moving point.
  • Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
  • The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
  • The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
  • By definition PM = PS

=> MP2 = PS2 

  • So, (a + x)2 = (x - a)2 + y2.
  • Hence, we can get the equation of horizontal parabola as y2 = 4ax.

For vertical parabola

  • Let us consider
  • Origin (0,0) as the parabola's vertex A
  1. Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
  2. P(x,y) as any moving point
  • Let us now draw a perpendicular SZ from S to the directrix.
  • Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
  • The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
  • By definition PM = PS

=> MP2 = PS2

So, (b + y)2 = (y - b)2 + x2

  • As a result, the vertical parabola equation is x2= 4by.