Q.2. Let one root of the quadratic equation in x: be twice the other. Then the length of the latus

Q: What is the correct answer for JEE Main Mathematics Q2?

The correct answer is D. Sol. Let the roots be $\alpha$ and $2\alpha$. Sum of roots: $\alpha + 2\alpha = 3\alpha = -\frac{9(k-1)}{k^2-15k+27}$ (1) Product of roots: $\alpha \cdot 2\alpha = 2\alpha^2 = \frac{18}{k^2-15k+27}$ (2) From (1) and (2): $3\alpha = -\frac{9(k-1)}{k^2-15k+27}$ and $2\alpha^2 = \frac{18}{k^2-15k+27}$. Divide (1) by (2): $\frac{3\alpha}{2\alpha^2} = \frac{-9(k-1)}{18} \Rightarrow \frac{3}{2\alpha} =

Q: Q.2. Let one root of the quadratic equation in x: be twice the other. Then the length of the latus rectum of the parabola is equal to:

The correct answer is D. Sol. Let the roots be $\alpha$ and $2\alpha$. Sum of roots: $\alpha + 2\alpha = 3\alpha = -\frac{9(k-1)}{k^2-15k+27}$ (1) Product of roots: $\alpha \cdot 2\alpha = 2\alpha^2 = \frac{18}{k^2-15k+27}$ (2) From (1) and (2): $3\alpha = -\frac{9(k-1)}{k^2-15k+27}$ and $2\alpha^2 = \frac{18}{k^2-15k+27}$. Divide (1) by (2): $\frac{3\alpha}{2\alpha^2} = \frac{-9(k-1)}{18} \Rightarrow \frac{3}{2\alpha} =

Question

Q.2. Let one root of the quadratic equation in x:  be twice the other. Then the length of the latus rectum of the parabola  is equal to:

FirstInTest · Accepted Answer

Sol. Let the roots be $\alpha$ and $2\alpha$. Sum of roots: $\alpha + 2\alpha = 3\alpha = -\frac{9(k-1)}{k^2-15k+27}$ (1) Product of roots: $\alpha \cdot 2\alpha = 2\alpha^2 = \frac{18}{k^2-15k+27}$ (2) From (1) and (2): $3\alpha = -\frac{9(k-1)}{k^2-15k+27}$ and $2\alpha^2 = \frac{18}{k^2-15k+27}$. Divide (1) by (2): $\frac{3\alpha}{2\alpha^2} = \frac{-9(k-1)}{18} \Rightarrow \frac{3}{2\alpha} = -\frac{k-1}{2} \Rightarrow \alpha = -\frac{3}{k-1}$. Substitute $\alpha$ into (2): $2\left(-\frac{3}{k-1}\right)^2 = \frac{18}{k^2-15k+27} \Rightarrow 2\frac{9}{(k-1)^2} = \frac{18}{k^2-15k+27} \Rightarrow \frac{1}{(k-1)^2} = \frac{1}{k^2-15k+27} \Rightarrow (k-1)^2 = k^2-15k+27 \Rightarrow k^2-2k+1 = k^2-15k+27 \Rightarrow -2k+1 = -15k+27 \Rightarrow 13k = 26 \Rightarrow k=2$. The parabola is $y^2 = 6kx = 6(2)x = 12x$. For a parabola $y^2 = 4ax$, the latus rectum is $4a$. Here, $4a = 12$. So, the length of the latus rectum is 12.

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