Question:
The minimum distance of a point on the curve
\[
y=x^2-4
\]
from the origin is
The minimum distance of a point on the curve
\[
y=x^2-4
\]
from the origin is
Show Hint
To find the minimum distance from the origin to a curve, minimize
\[
D^2=x^2+y^2
\]
instead of \(D\), because both have the same minimum point and \(D^2\) is easier to differentiate.
Updated On: Jun 26, 2026
- \(\frac{\sqrt{15}}{2}\)
- \(\frac{\sqrt{19}}{2}\)
- \(\sqrt{\frac{15}{2}}\)
- \(\sqrt{\frac{19}{2}}\)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Write the distance from origin.
Let a point on the curve \[ y=x^2-4 \] be \[ P(x,y). \] Since \[ y=x^2-4, \] we can write \[ P=(x,x^2-4). \] The distance of \(P\) from the origin \((0,0)\) is \[ D=\sqrt{x^2+y^2}. \] Substituting \[ y=x^2-4, \] we get \[ D=\sqrt{x^2+(x^2-4)^2}. \]
Step 2: Minimize the square of the distance.
To minimize \(D\), it is enough to minimize \(D^2\).
Let \[ D^2=x^2+(x^2-4)^2. \] Expanding, \[ D^2=x^2+x^4-8x^2+16. \] \[ D^2=x^4-7x^2+16. \] Let \[ F(x)=x^4-7x^2+16. \]
Step 3: Differentiate and find critical points.
Differentiate \(F(x)\): \[ F'(x)=4x^3-14x \] \[ F'(x)=2x(2x^2-7). \] For minimum or maximum, \[ F'(x)=0. \] So, \[ 2x(2x^2-7)=0. \] Hence, \[ x=0 \] or \[ 2x^2-7=0. \] Thus, \[ x^2=\frac{7}{2}. \]
Step 4: Check the minimum value.
When \[ x=0, \] we get \[ D^2=0^4-7(0)^2+16=16. \] When \[ x^2=\frac{7}{2}, \] we get \[ D^2=x^4-7x^2+16. \] Since \[ x^4=\left(\frac{7}{2}\right)^2=\frac{49}{4}, \] we get \[ D^2=\frac{49}{4}-7\cdot \frac{7}{2}+16. \] \[ D^2=\frac{49}{4}-\frac{49}{2}+16. \] Taking LCM \(4\), \[ D^2=\frac{49-98+64}{4}. \] \[ D^2=\frac{15}{4}. \] Therefore, \[ D=\sqrt{\frac{15}{4}} \] \[ D=\frac{\sqrt{15}}{2}. \]
Step 5: Final conclusion.
Hence, the minimum distance from the origin is \[ \boxed{\frac{\sqrt{15}}{2}} \]
Let a point on the curve \[ y=x^2-4 \] be \[ P(x,y). \] Since \[ y=x^2-4, \] we can write \[ P=(x,x^2-4). \] The distance of \(P\) from the origin \((0,0)\) is \[ D=\sqrt{x^2+y^2}. \] Substituting \[ y=x^2-4, \] we get \[ D=\sqrt{x^2+(x^2-4)^2}. \]
Step 2: Minimize the square of the distance.
To minimize \(D\), it is enough to minimize \(D^2\).
Let \[ D^2=x^2+(x^2-4)^2. \] Expanding, \[ D^2=x^2+x^4-8x^2+16. \] \[ D^2=x^4-7x^2+16. \] Let \[ F(x)=x^4-7x^2+16. \]
Step 3: Differentiate and find critical points.
Differentiate \(F(x)\): \[ F'(x)=4x^3-14x \] \[ F'(x)=2x(2x^2-7). \] For minimum or maximum, \[ F'(x)=0. \] So, \[ 2x(2x^2-7)=0. \] Hence, \[ x=0 \] or \[ 2x^2-7=0. \] Thus, \[ x^2=\frac{7}{2}. \]
Step 4: Check the minimum value.
When \[ x=0, \] we get \[ D^2=0^4-7(0)^2+16=16. \] When \[ x^2=\frac{7}{2}, \] we get \[ D^2=x^4-7x^2+16. \] Since \[ x^4=\left(\frac{7}{2}\right)^2=\frac{49}{4}, \] we get \[ D^2=\frac{49}{4}-7\cdot \frac{7}{2}+16. \] \[ D^2=\frac{49}{4}-\frac{49}{2}+16. \] Taking LCM \(4\), \[ D^2=\frac{49-98+64}{4}. \] \[ D^2=\frac{15}{4}. \] Therefore, \[ D=\sqrt{\frac{15}{4}} \] \[ D=\frac{\sqrt{15}}{2}. \]
Step 5: Final conclusion.
Hence, the minimum distance from the origin is \[ \boxed{\frac{\sqrt{15}}{2}} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Application of derivatives Questions
- If the normal drawn at a point $ P $ on the curve $ 3y = 6x - 5x^3 $ passes through the origin $(0, 0)$, then the positive integral value of the abscissa of point $ P $ is:
- The slope of the normal to the curve $y = 2x^2 + 3\sin x$ at $x = 0$ is:
- A balloon, which always remains spherical, has a variable radius. The rate at which its volume is increasing with respect to its radius $r$ when $r = 5$ cm is:
- The minimum value of the function $f(x) = x^2 + \frac{250}{x}$ for $x > 0$ is:
- If the length of the tangent at a point on the parabola $y^{2}=4ax$ is $4a\sqrt{5}$, then the length of the sub-normal at that point is
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions