Q87 / 150 · MHT CET 2026 (13 April Shift 1) · Mathematics
A point on the parabola $y^2 = \dfrac{36}{5}x$ at which the ordinate increases at thrice the rate of the abscissa is .....
- A$\left(\dfrac{6}{5}, \dfrac{1}{5}\right)$
- Correct answerB$\left(\dfrac{1}{5}, \dfrac{6}{5}\right)$
- C$(5, 6)$
- D$\left(\dfrac{3}{5}, \dfrac{9}{5}\right)$
Solution
Given the equation of the parabola:
$y^2 = \dfrac{36}{5}x$
Differentiating both sides with respect to time $t$:
$2y \dfrac{dy}{dt} = \dfrac{36}{5} \dfrac{dx}{dt}$
It is given that the ordinate increases at thrice the rate of the abscissa, so:
$\dfrac{dy}{dt} = 3 \dfrac{dx}{dt}$
Substituting this into the differentiated equation:
$2y \left(3 \dfrac{dx}{dt}\right) = \dfrac{36}{5} \dfrac{dx}{dt}$
$6y = \dfrac{36}{5}$
$y = \dfrac{6}{5}$
Substituting $y = \dfrac{6}{5}$ into the equation of the parabola to find $x$:
$\left(\dfrac{6}{5}\right)^2 = \dfrac{36}{5}x$
$\dfrac{36}{25} = \dfrac{36}{5}x$
$x = \dfrac{1}{5}$
The required point is $\left(\dfrac{1}{5}, \dfrac{6}{5}\right)$.
$y^2 = \dfrac{36}{5}x$
Differentiating both sides with respect to time $t$:
$2y \dfrac{dy}{dt} = \dfrac{36}{5} \dfrac{dx}{dt}$
It is given that the ordinate increases at thrice the rate of the abscissa, so:
$\dfrac{dy}{dt} = 3 \dfrac{dx}{dt}$
Substituting this into the differentiated equation:
$2y \left(3 \dfrac{dx}{dt}\right) = \dfrac{36}{5} \dfrac{dx}{dt}$
$6y = \dfrac{36}{5}$
$y = \dfrac{6}{5}$
Substituting $y = \dfrac{6}{5}$ into the equation of the parabola to find $x$:
$\left(\dfrac{6}{5}\right)^2 = \dfrac{36}{5}x$
$\dfrac{36}{25} = \dfrac{36}{5}x$
$x = \dfrac{1}{5}$
The required point is $\left(\dfrac{1}{5}, \dfrac{6}{5}\right)$.
Hence, the correct answer is option: 'B'.