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Q.
Minimum distance between the curves $y^{2}=x-1$ and $x^{2}=y-1$ is equal to
NTA AbhyasNTA Abhyas 2022Application of Derivatives
Solution:
General point on the curve $y^{2}=x-1$ is $\left(\right. t_{1}^{2} + 1 , \, \, t_{1} \left.\right)$ and the general point on the curve $x^{2}=y-1$ is $\left(\right. t_{2} , \, \, t_{2}^{2} + 1 \left.\right) .$ Since both the curves are symmetrical about the line $y=x$ , for the nearest point on the curve $y^{2}=x-1$ from the line $y=x$
Let, $ \, Q\left(t_{1}^{2} + 1, t_{1} \, \right)$
$P\left(\right.t_{2},t_{2}^{2}+1\left.\right)$
slopes of the tangents are same
$\frac{1}{2 t_{1}} = 2 \cdot t_{2}$
$t_{1}t_{2}=\frac{1}{4} \, \, $ ......(1)
Since $PQ$ is perpendicular to line $x=y$
$\frac{t_{1} - t_{2}^{2} - 1}{t_{1}^{2} + 1 - t_{2}}=-1 \, \, \, $ ......(2)
Solving (1) & (2), we get $t_{1}=t_{2}=\frac{1}{2}$
$P\left(\frac{1}{2} , \frac{5}{4}\right),Q\left(\frac{5}{4} , \frac{1}{2}\right)$
Therefore, $PQ=\frac{3 \sqrt{2}}{4}$ units