Let one root of the quadratic equation in x : (k^2-15k+27)x^2+9(k-1)x+18=0 be twice the ot…

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:

Accepted Answer

Let the roots of the quadratic equation

$({k}^{2}-15k+27){x}^{2}+9(k-1)x+18=0$

be $\alpha $ and $\beta $.

Given that one root is twice the other, let

$\alpha =2\beta $

So the roots are in the ratio $2:1$.

For a quadratic equation $a{x}^{2}+bx+c=0$:

$\alpha +\beta =-\frac{b}{a},\, \alpha \beta =\frac{c}{a}$

Here,

$a={k}^{2}-15k+27,\, b=9(k-1),\, c=18$

So,

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

and

$\alpha \beta =\frac{18}{{k}^{2}-15k+27}$

Now let the roots be $2t$ and $t$.

Then,

$\alpha +\beta =3t,\, \alpha \beta =2{t}^{2}$

Using the identity:

$(\alpha +\beta {)}^{2}=9{t}^{2}$

and

$\alpha \beta =2{t}^{2}$

So,

$(\alpha +\beta {)}^{2}=\frac{9}{2}\alpha \beta $

Now substitute the values from the quadratic:

${(\frac{-9(k-1)}{{k}^{2}-15k+27})}^{2}=\frac{9}{2}\cdot \frac{18}{{k}^{2}-15k+27}$

Simplify:

$\frac{81(k-1{)}^{2}}{({k}^{2}-15k+27{)}^{2}}=\frac{81}{{k}^{2}-15k+27}$

Cancel $81$:

$\frac{(k-1{)}^{2}}{({k}^{2}-15k+27{)}^{2}}=\frac{1}{{k}^{2}-15k+27}$

Cross-multiply:

$(k-1{)}^{2}={k}^{2}-15k+27$

Now expand:

${k}^{2}-2k+1={k}^{2}-15k+27$

Cancel ${k}^{2}$:

$-2k+1=-15k+27$

$13k=26$

$k=2$

Now the parabola is

${y}^{2}=6kx$

Substitute $k=2$:

${y}^{2}=12x$

Compare with standard form:

${y}^{2}=4ax$

So,

$4a=12\, \implies \, a=3$

Length of latus rectum of the parabola ${y}^{2}=4ax$ is

$4a$

Therefore,

$4a=12$

Hence, the required length is

$12$

So, the correct option is:

Solutions · 1