The point on the curve \(y^2 = 8x\) for which the abscissa and ordinate change at the same rate, is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (2, 4)
Let x and y change with time t. Differentiating \(y^2 = 8x\) with respect to t:
\(\displaystyle 2y\frac{dy}{dt} = 8\frac{dx}{dt}\)
The abscissa and ordinate change at the same rate when \(\displaystyle\frac{dy}{dt} = \frac{dx}{dt}\):
\(2y = 8\)
\(y = 4\)
\(\displaystyle x = \frac{y^2}{8} = \frac{16}{8} = 2\)
So the point is (2, 4).
Why the other options are incorrect
Option (1) (2, -4) — Incorrect. Here \(2y = -8\), so \(\frac{dy}{dt} = -\frac{dx}{dt}\): the ordinate decreases as fast as the abscissa increases. The rates are equal in size but opposite in sign, so they are not the same.
Option (3) \((1,2\sqrt{2})\) — Incorrect. Here \(2y = 4\sqrt{2} \ne 8\), so \(\frac{dy}{dt} = \sqrt{2}\,\frac{dx}{dt}\).
Option (4) \((1,-2\sqrt{2})\) — Incorrect. Here \(2y = -4\sqrt{2}\), so \(\frac{dy}{dt} = -\sqrt{2}\,\frac{dx}{dt}\).