JEE Main 202311 AprilMorning Shift
- Exam
- JEE Main
- Year
- 2023
- Date
- 11 April
- Shift
- Morning Shift
- Subject
- Mathematics
- Chapter
- Hyperbola
JEE Main 2023 · 11 April · Morning Shift · Mathematics · Hyperbola
Let R be a rectangle given by the lines x = 0, x = 2 , y = 0 and y...
Let x = 0, x = 2y = 0y = 5 and and , be such that the line segment divides the area of the rectangle in the ratio . Then, the mid-point of lies on a![Let R be a rectangle given by the lines x = 0, x = 2 , y = 0 and y = 5 . Let A(α, 0) and B(0, β), α ∈ [0, 2] and β ∈](/api/proxy-image?clean=1&b=aHR0cHM6Ly9jZG4tcXVlc3Rpb24tcG9vbC5nZXRtYXJrcy5hcHAvcHlxL2plZV9tYWluLzJjNTE3YzE3LTg3NmItNDkwNy05OWY2LWJiNWVhM2ExOTA2YS5wbmc)
Options
- Astraight line
- Bparabola
- Chyperbola
- Dcircle
Correct answer
C. hyperbola
Step-by-step solution
Given,
be a rectangle given by the lines ,
And point A \left(\alpha, 0\right), \alpha \in \left[0, 2\right],
Now plotting the diagram we get,
Now given that line segment divides the ratio of area in , we get
\dfrac{10 - \dfrac{1}{2}\alpha\beta}{\dfrac{1}{2}\alpha\beta} = \dfrac{4}{1}</p><p>⇒ 20 - \alpha\beta = 4\alpha\beta⇒ \alpha\beta = 4 . . . . . . \left(1\right)h = \dfrac{\alpha}{2}\text{} \beta =\dfrac{k}{2} we get,
which is a equation of rectangular hyperbola.