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

52. Find the equation of the tangent line to the curve y = x² – 2x +7 which is

(a) Parallel to the line 2x – y + 9 = 0 (b) Perpendicular to the line 5y – 15x = 13

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Vishal Baghel | Contributor-Level 10

4 months ago

The eqⁿof the given curve is $y = x^{2} - 2 x + 7$

Slope of tangent, $\frac{d y}{d x} = 2 x - 2$

(a) The line $2 x - y + 9 = 0 \Rightarrow y = 2 x + 9$ compared to $y = m x + c$ gives,

Slope of line = 2.

If the tangent of the curve is parallel to the line

$\frac{d y}{d x} =slope of line$

$\to 2 x - 2 = 2$

$\to x = \frac{4}{2} \Rightarrow x = 2$

When $x = 2, y = {(2)}^{2} - 2 (2) + 7 = 7$

Hence, the point of contact of the tangent is (2, 7)

The eqⁿ of tangent is $y - 7 = 2 (x - 2)$

$\Rightarrow y - 7 = 2 x - 4$

$\to 2 x - y + 3 = 0$

(b) The line $5 y - 15 x = 13 \to y = \frac{15}{5} x + \frac{13}{5} \Rightarrow y = 3 x + \frac{13}{5}$

compared to $y = m x + c$ gives

slope of line = 3

As the tangent to the curve is ⊥ to the line.

$\frac{d y}{d x} = -1/slope opf line$

$\Rightarrow 2 x - 2 = - \frac{1}{3}$

$\Rightarrow 6 x - 6 = - 1$

$\to 6 x = 5$

$\to x = \frac{5}{6}$

When $x = \frac{5}{6}$ we get $y = {(\frac{5}{6})}^{2} - 2 \times \frac{5}{6} + 7$

$= \frac{25}{36} - \frac{5}{3} + 7$

$= \frac{25 - 60 + 252}{36} = \frac{217}{36}$

Hence, the point of contact of the tangent is $(\frac{5}{6}, \frac{217}{36})$

And eqⁿ of the tangent is

$y - \frac{217}{36} = \frac{- 1}{3} (x - \frac{5}{6})$

$\Rightarrow 3 y - \frac{217}{12} = - x + \frac{5}{6}$

$\Rightarrow x + 3 y - \frac{217}{12} - \frac{5}{6} = 0$

$\Rightarrow x + 3 y - \frac{217 - 10}{12} = 0$

$\Rightarrow x + 3 y - \frac{227}{12} = 0$

$\Rightarrow 12 x + 36 y - 227 = 0 .$

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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|

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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 $\sqrt[]{3} r .$

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52. Find the equation of the tangent line to the curve y = x² – 2x +7 which is

(a) Parallel to the line 2x – y + 9 = 0 (b) Perpendicular to the line 5y – 15x = 13

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52. Find the equation of the tangent line to the curve y = x2 – 2x +7 which is (a) Parallel to the line 2x – y + 9 = 0 (b) Perpendicular to the line 5y – 15x = 13

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52. Find the equation of the tangent line to the curve y = x² – 2x +7 which is

(a) Parallel to the line 2x – y + 9 = 0 (b) Perpendicular to the line 5y – 15x = 13