SATHEE: Application of Derivatives 4 Question 6

Application of Derivatives 4 Question 6

####7. The shortest distance between the line $y = x$ and the curve $y^{2} = x - 2$ is

(a) 2

(b) $\frac{7}{8}$

(d) $\frac{11}{4 \sqrt{2}}$

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Answer:

Correct Answer: 7. (c)

Solution:

Given equation of curve is

$y^{2} = x - 2$

and the equation of line is

Consider a point $P (t^{2} + 2, t)$ on parabola (i).

For the shortest distance between curve (i) and line (ii), the line $P M$ should be perpendicular to line (ii) and parabola (i), i.e. tangent at $P$ should be parallel to $y = x$ . ${∴ \frac{d y}{d x} |}_{at point P} =$ Slope of tangent at point $P$ to curve (i) $= 1 [∵$ tangent is paralle $]$

${\Rightarrow \frac{1}{2 y} |}_{P} = 1$

to line $y = x]$

[differentiating the curve (i), we get $2 y \frac{d y}{d x} = 1$ ]

$\Rightarrow \frac{1}{2 t} = 1 \Rightarrow t = \frac{1}{2} [∵ P (x, y) = P (t^{2} + 2, t)]$

So, the point $P$ is $\frac{9}{4}, \frac{1}{2}$ .

Now, minimum distance $= P M = \frac{\frac{9}{4} - \frac{1}{2}}{\sqrt{2}}$

$[∵$ distance of a point $P (x_{1}, y_{1})$ from a line

$\begin{matrix} a x + b y + c = 0 is \frac{| a x_{1} + b y_{1} + c |}{\sqrt{a^{2} + b^{2}}} \\ = \frac{7}{4 \sqrt{2}} units \end{matrix}$

Application of Derivatives 4 Question 6

Answer:

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Application of Derivatives 4 Question 6

Answer:

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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