Question

If parabola $y^2=\lambda x$ and $25\left[\right(x-3{)}^2+(y+2{)}^2]=(3x-4y-2{)}^2$ are equal , then value of $\lambda$ is

• A. $9$
• B. $3$
• C. $7$
• D. $6$

Answer 1

Answer: (D)

$6$

Two parabolas are equal if the length of their latus rectum are equal.
Length of the latus rectum of $y^2=\lambda x$ is $\lambda$.
The equation of the second parabola is
$25\left\{\right(x-3{)}^2+(y+2{)}^2\}=(3x-4y-2{)}^2$
$\Rightarrow \sqrt{(x-3{)}^2+(y+2{)}^2}=\frac{|3x-4y-2|}{\sqrt{3^2+4^2}}$
Here focus is $(3,-2)$ and equation of the directrix is $3x-4y-2=0$
Therefore, length of the latus rectum $=2\times$ distance between focus and directrix
$=2\left|\frac{3\times 3-4\times (-2)-2}{\sqrt{3^2+(-4{)}^2}}\right|=6$
Thus, the two parabolas are equal, if $\lambda =6$.