Question

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

• A. $9$
• B. $3$
• C. $7$
• D. $6$

Answer 1

The correct option is D $6$

Let us recall that two parabolas are equal if the length of their latus rectum are equal so, the length of latus rectum of $y=\lambda x$ is $\lambda$ The eqn of second parabola is $25(x-3{)}^2+(y+2{)}^2)=(3x-4y-z{)}^2$

$\Rightarrow \sqrt{(x-3{)}^2+(y+2{)}^2}=\frac{3x-4y-2}{\sqrt{3^2+4^2}}$, So clearly it represents a parabola having focus at $(3,-2)$ and eqn direction is $3x-4y=2\Rightarrow 3x+4y-2=0$

So length of latus rectum $=2$ (distance between focus & direction)

$=2\left|\frac{3\times 3-4x(-2)-2}{\sqrt{3^2+(-4{)}^2}}\right|$

$\lambda =6$ ( since two parabola are equal)