SATHEE: Parabola 2 Question 20

Parabola 2 Question 20

20. Find the shortest distance of the point $(0, c)$ from the parabola $y = x^{2}$ , where $0 \leq c \leq 5$ .

$(1982, 2 M)$

Integer Answer Type Question

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Solution:

Let the point $Q (x, x^{2})$ on $x^{2} = y$ whose distance from $(0, c)$ is minimum.

Now, $P Q^{2} = x^{2} + {(x^{2} - c)}^{2}$

$Let \begin{aligned} f (x) & = x^{2} + {(x^{2} - c)}^{2} \\ f^{'} (x) & = 2 x + 2 (x^{2} - c) \cdot 2 x \\ = 2 x (1 + 2 x^{2} - 2 c) = 4 x x^{2} - c + \frac{1}{2} \end{aligned}$

$= 4 x x - \sqrt{c - \frac{1}{2}} x + \sqrt{c - \frac{1}{2}}, when c > \frac{1}{2}$

For maxima, put $f^{'} (x) = 0$

$4 x x^{2} - c + \frac{1}{2} = 0 \Rightarrow x = 0, x = \pm \sqrt{c - \frac{1}{2}}$

Now,

$f^{''} (x) = 4 x^{2} - c + \frac{1}{2} + 4 x [2 x]$

$x = \pm \sqrt{c - \frac{1}{2}}$

$f^{''} (x) \geq 0$ . $∴ f (x)$ is minimum.

Hence, minimum value of $f (x) \neq P Q ∣$

$\begin{aligned} = \sqrt{c - \frac{1}{2} + c - \frac{1}{2} - c^{2}} = \sqrt{c - \frac{1}{4}}, \frac{1}{2} \leq c \leq 5 \end{aligned}$

Parabola 2 Question 20

20. Find the shortest distance of the point $(0, c)$ from the parabola $y = x^{2}$ , where $0 \leq c \leq 5$ .

Integer Answer Type Question

Solution:

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Parabola 2 Question 20

20. Find the shortest distance of the point (0,c) from the parabola y=x2, where 0≤c≤5.

Integer Answer Type Question

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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20. Find the shortest distance of the point $(0, c)$ from the parabola $y = x^{2}$ , where $0 \leq c \leq 5$ .