Question

The parabola $y^2=\lambda x$ and $25\left[\right(x-{3)}^2+(y-{2)}^2]=(3x-4y-{2)}^2$ are equal, if $\lambda$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $6$

Answer 1

The correct option is D $6$

Two parabolas are equal if the length of their latus rectum are equal.

So, length of latus rectum of $y^2=\lambda nis\lambda ⟶\left(1\right)$

The equation for second parabola is

$25\{{(n-3)}^2+{(y+2)}^3\}={(3n-4y-2)}^{22}\Rightarrow \sqrt{{(n-3)}^2+{(y+2)}^2}=\frac{|3n-4y-2|}{\sqrt{3^2+4^2}}$

Clearly, it represents parabola having focus at (3,-2) and equation of the directrix as 3n-4y-2=0.

Length of the latus rectum =$2\times \left(distancebetweenfocusanddirectrix\right)2\left|\frac{3\times 3-4\times (-2)-2}{\sqrt{3^2+{(-4)}^2}}\right|=6⟶\left(II\right)$

From (I) and (II),

$\lambda =6$