Application of Derivatives
The shortest distance between the line $$y-x =1$$ and the curve $$\mathrm{x}=\mathrm{y}^{2}$$ is
The shortest distance between line $$y-x =1$$ and curve $$\mathrm{x}=\mathrm{y}^{2}$$ is
Let $$P=(y^2,y)$$ be a point on the curve.
The perpendicular distance from $$P$$ to $$x-y+1=0$$ is $$\displaystyle \dfrac{|y^2-y+1|}{\sqrt{2}}$$
Now, the discriminant for the expression $$y^2-y+1 $$ is negative and the co-efficient of $$y^2$$ is positive. Hence, the expression $$y^2-y+1 $$ is always positive.
For the minimum value of distance, $$ y =\dfrac{1}{2} $$.
Hence,
Minimum distance $$ = \dfrac{\frac{3}{4} }{\sqrt {2}} =\dfrac{3\sqrt{2}}{8}$$
The shortest distance between the line $$y-x =1$$ and the curve $$\mathrm{x}=\mathrm{y}^{2}$$ is
The point $$(0, 5)$$ is closer to the curve $$\displaystyle { x }^{ 2 }=2y$$ at:
The shortest distance between the parabolas $${ y }^{ 2 }=4x$$ and $${ y }^{ 2 }=2x-6$$ is
The coordinates of the point on the parabola $$\displaystyle y^{2}=8x,$$ which is at minimum distance from the circle $$\displaystyle x^{2}+\left ( y+6 \right )^{2}=1$$ are
On the parabola $$y={ x }^{ 2 },$$ the point at a least distance from the straight line $$y=2x-4$$ is
Minimum distance between the curves $$y^2=4x\;\&\;x^2+y^2-12x+31=0$$ is
The number of point in the rectangle $$\{(x,y);$$ $$−12\le x \le12$$ and $$−3\le y\le3\}$$ which lie on the curve $$y=x+\sin x$$ and at which in the tangent to the curve is parallel to the $$x-axis$$ is
If d is the minimum distance between the curves $$f(x)=e^{x}$$ and $$g(x) = log_{e}x,$$ then the value of $$d^{6}$$ is
Find the shortest distance of the point (0,c) from the parabola $$y = x^{2}$$, where $$0\leq c \leq 5$$.