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

In this question, we are given a quadratic equation $(k^2 - 15k + 27)x^2 + 9(k - 1)x + 18 = 0$ where one root is twice the other root. Our goal is to find the value of $k$ to determine the length of the latus rectum of the parabola $y^2 = 6kx$.

Let the roots of the given quadratic equation be $\alpha$ and $2\alpha$.
For a quadratic equation $Ax^2 + Bx + C = 0$, the sum of roots is $-B/A$ and the product of roots is $C/A$.
Here, $A = (k^2 - 15k + 27)$, $B = 9(k-1)$, and $C = 18$.

1. Sum of roots:
$\alpha + 2\alpha = 3\alpha = \frac{-9(k-1)}{k^2 - 15k + 27}$
Dividing by 3, we get:
$\alpha = \frac{-3(k-1)}{k^2 - 15k + 27}$ ... (i)

2. Product of roots:
$\alpha \times 2\alpha = 2\alpha^2 = \frac{18}{k^2 - 15k + 27}$
Dividing by 2, we get:
$\alpha^2 = \frac{9}{k^2 - 15k + 27}$ ... (ii)

3. Substituting the value of $\alpha$ from (i) into (ii):
$\left( \frac{-3(k-1)}{k^2 - 15k + 27} \right)^2 = \frac{9}{k^2 - 15k + 27}$
$\frac{9(k-1)^2}{(k^2 - 15k + 27)^2} = \frac{9}{k^2 - 15k + 27}$

Assuming $k^2 - 15k + 27 \neq 0$, we can cancel $9$ and one factor of $(k^2 - 15k + 27)$ from both sides:
$(k-1)^2 = k^2 - 15k + 27$
$k^2 - 2k + 1 = k^2 - 15k + 27$
$13k = 26$
$k = 2$

4. The equation of the parabola is $y^2 = 6kx$.
Substituting $k=2$, we get $y^2 = 6(2)x$, which is $y^2 = 12x$.
The standard form of a parabola is $y^2 = 4ax$, where the length of the latus rectum is $4a$.
Comparing $y^2 = 12x$ with $y^2 = 4ax$, we find the length of the latus rectum to be 12.