Answer:
Let the required point be (x, y). Given, \[\frac{dy}{dt}=\frac{dx}{dt}\] and \[{{y}^{2}}=8x\] On differentiating both sides of Eq. (ii) w.r.t.t, (we get \[2y\frac{dy}{dt}=8\cdot \frac{dx}{dt}\] \[\Rightarrow \] \[2y=8\] [from Eq. (i)] \[\Rightarrow \] \[y=4\] On putting y = 4 in Eq. (ii), we get \[8x=16\,\,\,\Rightarrow \,\,\,x=2\] Hence, the required point is (2, 4).
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