Q120 / 180 · COMEDK 2024 (Evening Shift) · Mathematics
In the parabola $y^2=4 a x$ the length of the latus rectum is 6 units and there is a chord passing through its vertex and the negative end of the latus rectum. Then the equation of the chord is
- Correct answerA$2 x+y=0$
- B$x-2 y=0$
- C$x+2 y=0$
- D$2 x-y=0$
Solution
The equation of the parabola is $y^2 = 4ax$. The length of the latus rectum is given as $4a = 6$, which implies $a = \dfrac{6}{4} = \dfrac{3}{2}$.
The coordinates of the vertex are $(0, 0)$. The endpoints of the latus rectum are $(a, 2a)$ and $(a, -2a)$. The negative end of the latus rectum is $(a, -2a)$.
Substituting $a = \dfrac{3}{2}$, the coordinates of the negative end of the latus rectum are $(\dfrac{3}{2}, -2 \times \dfrac{3}{2}) = (\dfrac{3}{2}, -3)$.
The chord passes through the vertex $(0, 0)$ and the point $(\dfrac{3}{2}, -3)$. The slope $m$ of this chord is $m = \dfrac{-3 - 0}{\dfrac{3}{2} - 0} = \dfrac{-3}{\dfrac{3}{2}} = -2$.
The equation of the line passing through the origin with slope $m = -2$ is $y = -2x$, which can be rewritten as $2x + y = 0$.
Answer: $2 x+y=0$
The coordinates of the vertex are $(0, 0)$. The endpoints of the latus rectum are $(a, 2a)$ and $(a, -2a)$. The negative end of the latus rectum is $(a, -2a)$.
Substituting $a = \dfrac{3}{2}$, the coordinates of the negative end of the latus rectum are $(\dfrac{3}{2}, -2 \times \dfrac{3}{2}) = (\dfrac{3}{2}, -3)$.
The chord passes through the vertex $(0, 0)$ and the point $(\dfrac{3}{2}, -3)$. The slope $m$ of this chord is $m = \dfrac{-3 - 0}{\dfrac{3}{2} - 0} = \dfrac{-3}{\dfrac{3}{2}} = -2$.
The equation of the line passing through the origin with slope $m = -2$ is $y = -2x$, which can be rewritten as $2x + y = 0$.
Answer: $2 x+y=0$
Hence, the correct answer is option: 'A'.