47. Find the equation of all lines having slope – 1 that are tangents to the curve <math><mrow><mi>y</mi><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mi>x</mi><mo>−</mo><mn>1</mn></mrow></mfrac><mo>,</mo><mi>x</mi><mo>≠</mo><mn>1</mn></mrow></math>

The given eqn of curve is y=1x−1Slope of tangent to the given curve is dydx=−1(x−1)2Given that, slope of tangent = 1.→−1(x−1)2=−1.⇒(x−1)2=1.⇒ x−1=±1→x=1±1.ie, X=1+1 or x=1−1⇒x=2 or x=0When x=2,y=12−1=1and when x=0, y=10−1=−1.Hence, the point of contact of the tangents are (2,1)and(0,−1)The reqd. eqn of line are y−1=(−1)(x−2) {?y−y0=m(x−x0) eqn of line and y−(−1)=(−1)(x−0).⇒y−1=−x+2 and y+1=−x→x+y−3=0 and x+y+1=0.

Ask & Answer Question

Ncert Solutions Maths class 12th Maths Class 12th Application of Derivatives

47. Find the equation of all lines having slope – 1 that are tangents to the curve $y = \frac{1}{x - 1}, x \neq 1$

2 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

The given eqⁿ of curve is $y = \frac{1}{x - 1}$

Slope of tangent to the given curve is $\frac{d y}{d x} = \frac{- 1}{{(x - 1)}^{2}}$

Given that, slope of tangent = 1.

$\to \frac{- 1}{{(x - 1)}^{2}} = - 1 .$

$\Rightarrow {(x - 1)}^{2} = 1 .$

$\Rightarrow x - 1 = \pm 1$

$\to x = 1 \pm 1 .$

ie, X=1+1 or $x = 1 - 1$

$\Rightarrow x = 2$ or $x = 0$

When $x = 2, y = \frac{1}{2 - 1} = 1$

and when $x = 0, y = \frac{1}{0 - 1} = - 1 .$

Hence, the point of contact of the tangents are $(2, 1)and (0, - 1)$

The reqd. eqⁿ of line are $y - 1 = (- 1) (x - 2) {\begin{matrix} ? y - y_{0} = m (x - x_{0}) & eqn^{} of line \end{matrix}$

and $y - (- 1) = (- 1) (x - 0) .$

$\Rightarrow y - 1 = - x + 2$ and $y + 1 = - x$

$\to x + y - 3 = 0$ and $x + y + 1 = 0 .$

0 Upvotes 0 Downvotes

Similar Questions for you

If the normal to the curve y(x) = ∫(from 2 to x) (2t² - 15t + 10)dt at a point (a, b) is parallel to the line x + 3y = -5, a > 1, then the value of |a + 6b| is equal to..........

Raj Pandey

y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
For tangent at (a, b), slope is m = dx/dy = 1 / (dy/dx) = 1 / (2a² - 15a + 10).
Given slope is -1/3.
2a² - 15a + 10 = -3
2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a + 7 = 0
(2a - 1) (a - 7) = 0
a = 1/2 or a = 7.
a = 1/2 Rejected as a > 1. So a = 7.
b = ∫? (2t² - 15t + 10)dt = [2t³/3 - 15t²/2 + 10t] from 0 to 7.
6b = [4t³ - 45t² + 60t] from 0 to 7 = 4 (7)³ - 45 (7)² + 60 (7) = 1372 - 2205 + 420 = -413.
|a + 6b| = |7 - 413| = |-406|

...more

If the tangent to the curve, y = f(x) = xlog? x, (x > 0) at a point (c, f(c)) is parallel to the line - segment joining the points (1,0) and (e, e) then c is equal to:

Raj Pandey

f' (c) = 1 + lnc = e/ (e-1)
lnc = e/ (e-1) - 1 = (e - (e-1)/ (e-1) = 1/ (e-1)
c = e^ (1/ (e-1)

The lengths of the sides of a triangle are 10 + x2, 10 + x2 and 20 – 2x2. If for x = k, the area of the triangle is maximum, then 3k2 is equal to:

Raj Pandey

$C D = \sqrt[]{{(10 + x^{2})}^{2} - {(10 - x^{2})}^{2}} = 2 \sqrt[]{10} | x |$

Area

$= \frac{1}{2} \times C D \times A B = \frac{1}{2} \times 2 \sqrt[]{10} | x | (20 - 2 x^{2})$

$\Rightarrow 10 - x^{2} = 2 x$

3x² = 10

x = k

3k² = 10

Let F₁(A,B,C) = $(A \land \sim B) \lor$ $[\sim C \land (A \lor B)] \lor \sim A a n d F_{2} (A, B) = (A \lor B) \lor (B \to \sim A)$ be two logical expressions. Then:

Vishal Baghel

By truth table

So F₁ (A, B, C) is not a tautology

Now again by truth table

So F₂ (A, B) be a tautology.

The triangle of maximum area that can be inscribed in a given circle of radius ‘r’ is:

Vishal Baghel

From option let it be isosceles where AB = AC then

$x = \sqrt[]{r^{2} - {(h - r)}^{2}}$

= $\sqrt[]{r^{2} - h^{2} - r^{2} + 2 r h}$

$x = \sqrt[]{2 h r - h^{2}} . . . . . . . . (i)$

Now ar $(Δ A B C) = Δ = \frac{1}{2} B C \times A L$

$Δ = \frac{1}{2} \times 2 \sqrt[]{2 h r - a^{2}} \times h$

then $x = \sqrt[]{2 \times \frac{3 r}{2} \times r - \frac{9 r^{2}}{4}} = \frac{\sqrt[]{3}}{2} r f r o m (i)$

$\Rightarrow B C = \sqrt[]{3} r$

So $A B = \sqrt[]{h^{2} + x^{2}} = \sqrt[]{\frac{9 r^{2}}{4} + \frac{3 r^{2}}{4}} = \sqrt[]{3} r$ .

Hence $Δ$ be equilateral having each side of length $\sqrt[]{3} r .$

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

47. Find the equation of all lines having slope – 1 that are tangents to the curve $y = \frac{1}{x - 1}, x \neq 1$

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

47. Find the equation of all lines having slope – 1 that are tangents to the curve y=1x−1,x≠1

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

47. Find the equation of all lines having slope – 1 that are tangents to the curve $y = \frac{1}{x - 1}, x \neq 1$