The coordinates of the points on the curve 4y | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let a point on the curve $4y = x^2$ be $(x, y)$, where $y = \frac{x^2}{4}$.The distance $D$ between $(x, \frac{x^2}{4})$ and $(0, 5)$ is given by

Vedclass · Accepted Answer

$(2\sqrt{3}, 3)$

Vedclass · Answer

(B) Let a point on the curve $4y = x^2$ be $(x, y)$, where $y = \frac{x^2}{4}$.The distance $D$ between $(x, \frac{x^2}{4})$ and $(0, 5)$ is given by $D^2 = (x - 0)^2 + (\frac{x^2}{4} - 5)^2$.Let $f(x) = D^2 = x^2 + \frac{x^4}{16} - \frac{5x^2}{2} + 25 = \frac{x^4}{16} - \frac{3x^2}{2} + 25$.To find the minimum distance, differentiate $f(x)$ with respect to $x$ and set it to $0$:$f'(x) = \frac{4x^3}{16} - 3x = \frac{x^3}{4} - 3x = 0$.$x(\frac{x^2}{4} - 3) = 0$, which gives $x = 0$ or $x^2 = 12 \implies x = \pm 2\sqrt{3}$.For $x = \pm 2\sqrt{3}$, $y = \frac{(\pm 2\sqrt{3})^2}{4} = \frac{12}{4} = 3$.Thus, the points are $(2\sqrt{3}, 3)$ and $(-2\sqrt{3}, 3)$.

The coordinates of the points on the curve $4y = x^2$ that are nearest to the point $(0, 5)$ are ...

Similar Questions

The maximum value of $f(x) = \frac{x}{4 + x + x^2}$ on the interval $[-1, 1]$ is:

The local maximum value of the function $f(x)=-(x-2)^3(x+2)^2$ is

The two parts of $100$ for which the sum of double of the first part and the square of the second part is minimum,are

The maximum value of $\frac{\log x}{x}$ in $(2, \infty)$ is

The difference between the local extreme values of the function $f(x) = 2x^3 - 15x^2 + 36x + 40$ is ......

Vedclass Test Series

Exam Paper Generator

Online Exam Module