Question

Let one root of the quadratic equation in $x$:
$(k^2 - 15k + 27)x^2 + 9(k - 1)x + 18 = 0$
be twice the other. Then the length of the latus rectum of the parabola $y^2 = 6kx$ is equal to:

• A. 4
• B. 6
• C. 8
• D. 12

Hint:

Let the roots be $\alpha$ and $2\alpha$. Use the sum and product of roots formulas to find $k$. The length of the latus rectum for $y^2 = 4ax$ is $4a$.

Answer 1

The Correct Option is D

To solve for the length of the latus rectum of $y^2 = 6kx$, we first need to identify the parameter $k$ from the relationship between the roots of the quadratic equation $(k^2 - 15k + 27)x^2 + 9(k - 1)x + 18 = 0$.

Let the roots be $\alpha$ and $\beta$. We are given $\beta = 2\alpha$.
We know that for a quadratic $Ax^2 + Bx + C = 0$:
$\alpha + \beta = -B/A$ and $\alpha\beta = C/A$.

Substituting $\beta = 2\alpha$:
$3\alpha = \frac{-9(k-1)}{k^2 - 15k + 27}$
$2\alpha^2 = \frac{18}{k^2 - 15k + 27}$

From the first equation, $\alpha = -\frac{3(k-1)}{k^2 - 15k + 27}$. Squaring this gives:
$\alpha^2 = \frac{9(k-1)^2}{(k^2 - 15k + 27)^2}$

Equating this to the value of $\alpha^2$ from the product of roots equation:
$\frac{9(k-1)^2}{(k^2 - 15k + 27)^2} = \frac{9}{k^2 - 15k + 27}$
This simplifies to $(k-1)^2 = k^2 - 15k + 27$.
Expanding the left side: $k^2 - 2k + 1 = k^2 - 15k + 27$.
Solving for $k$: $13k = 26 \implies k = 2$.

The parabola is $y^2 = 6kx = 12x$.
The latus rectum of a parabola $y^2 = 4ax$ is defined as the coefficient of $x$, which is $4a$.
Therefore, the length of the latus rectum is 12.