SATHEE: Application of Derivatives 4 Question 49

Application of Derivatives 4 Question 49

####52. What normal to the curve $y = x^{2}$ forms the shortest chord?

(1992, 6M)

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Answer:

Correct Answer: 52. $x = 0, y = 0$ 59. $\frac{u d}{\sqrt{v^{2} - u^{2}}}$

Solution:

Any point on the parabola $y = x^{2}$ is of the form $(t, t^{2})$ .

Now, $\frac{d y}{d x} = 2 x \Rightarrow \frac{d y}{d x}_{x = t} = 2 t$

Which is the slope of the tangent. So, the slope of the normal to $y = x^{2}$ at $A (t, t^{2})$ is $- 1 / 2 t$ .

Therefore, the equation of the normal to

$\begin{aligned} y & = x^{2} at A (t, t^{2}) is \\ y - t^{2} & = - \frac{1}{2 t} (x - t) \end{aligned}$

Suppose Eq. (i) meets the curve again at $B (t_{1}, t_{1}^{2})$ . Then, $t_{1}^{2} - t^{2} = - \frac{1}{2 t} (t_{1} - t)$

$\Rightarrow (t_{1} - t) (t_{1} + t) = - \frac{1}{2 t} (t_{1} - t)$

$\Rightarrow (t_{1} + t) = - \frac{1}{2 t}$

$\Rightarrow t_{1} = - t - \frac{1}{2 t}$

Therefore, length of chord,

$\begin{aligned} L & = A B^{2} = {(t - t_{1})}^{2} + {(t^{2} - t_{1}^{2})}^{2} \\ = {(t - t_{1})}^{2} + {(t - t_{1})}^{2} {(t + t_{1})}^{2} \\ = {(t - t_{1})}^{2} [1 + {(t + t_{1})}^{2}] \\ = t + t + {\frac{1}{2 t}}^{2} 1 + t - t - {\frac{1}{2 t}}^{2} \\ \Rightarrow L & = 2 t + {\frac{1}{2 t}}^{2} 1 + \frac{1}{4 t^{2}} = 4 t^{2} 1 + {\frac{1}{4 t^{2}}}^{3} \end{aligned}$

On differentiating w.r.t. $t$ , we get

$\begin{aligned} \frac{d L}{d t} & = 8 t 1 + {\frac{1}{4 t^{2}}}^{3} + 12 t^{2} 1 + {\frac{1}{4 t^{2}}}^{2} - \frac{2}{4 t^{3}} \\ = 2 1 + {\frac{1}{4 t^{2}}}^{2} 4 t 1 + \frac{1}{4 t^{2}} - \frac{3}{t} \\ = 2 1 + {\frac{1}{4 t^{2}}}^{2} 4 t - \frac{2}{t} = 4 1 + {\frac{1}{4 t^{2}}}_{2}^{2} 2 t - \frac{1}{t} \end{aligned}$

For maxima or minima, we must have $\frac{d L}{d t} = 0$

$\begin{aligned} \Rightarrow & 2 t - \frac{1}{t} & = 0 \Rightarrow \\ \Rightarrow & t & = \pm \frac{1}{\sqrt{2}} \end{aligned}$

Now, $\frac{d^{2} L}{d t^{2}} = 8 1 + \frac{1}{4 t^{2}} - \frac{1}{2 t^{3}} 2 t - \frac{1}{t}$

$\Rightarrow \frac{d^{2} L}{d t^{2}}_{t = \pm 1 / \sqrt{2}} = 0 + 4 1 + {\frac{1}{4 t^{2}}}^{2} 2 + \frac{1}{t^{2}}$

Therefore, $L$ is minimum, when $t = \pm 1 / \sqrt{2}$ . For $t = 1 / \sqrt{2}$ , point $A$ is $(1 / \sqrt{2}, 1 / 2)$ and point $B$ is $(- \sqrt{2}, 2)$ . When $t = - 1 / \sqrt{2}, A$ is $(- 1 / \sqrt{2}, 1 / 2), B$ is $(\sqrt{2}, 2)$ .

Again, when $t = 1 / \sqrt{2}$ , the equation of $A B$ is

$\begin{aligned} \frac{y - 2}{\frac{1}{2} - 2} & = \frac{x + \sqrt{2}}{\frac{1}{\sqrt{2}} + \sqrt{2}} \\ \Rightarrow (y - 2) \frac{1}{\sqrt{2}} + \sqrt{2} & = (x + \sqrt{2)} \frac{1}{2} - 2 \\ \Rightarrow \sqrt{2} x + 2 y - 2 & = 0 \end{aligned}$

and when $t = - 1 / \sqrt{2}$ , the equation of $A B$ is

$\begin{aligned} \frac{y - 2}{\frac{1}{2} - 2} & = \frac{x - \sqrt{2}}{- \frac{1}{\sqrt{2}} - \sqrt{2}} \\ \Rightarrow (y - 2) - \frac{1}{\sqrt{2}} - \sqrt{2} & = (x - \sqrt{2}) \frac{1}{2} - 2 \\ \Rightarrow \sqrt{2} x - 2 y + 2 & = 0 \end{aligned}$

Application of Derivatives 4 Question 49

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Application of Derivatives 4 Question 49

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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