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

**60. Prove that the curves x = y²and xy = k cut at right angles* if 8k² = 1.**

2 Views|Posted 9 months ago

Asked by Shiksha User

1 Answer

Sort by:

V

Answered by

Vishal Baghel | Contributor-Level 10

9 months ago

The given eqⁿ of the curves are

$x = y^{2}$ ________(1)

and $x y = x$ ___________(2)

Differentiating eqn (1) and (2) wrt 'x' we get,

$\frac{d y^{2}}{d x} = \frac{d x}{d x}$

$\to 2 y \frac{d y}{d x} = 1$

$\to \frac{d y}{d x} = \frac{1}{2 y}$ _________(3)

and $\frac{d}{d x} (x y) = \frac{d}{d x} (x)$

$\to x \frac{d y}{d x} + y = 0$

$\to \frac{d y}{d x} = - \frac{y}{x}$ _________(4)

Since the two curves cut each other at right angles we get,

$(\frac{1}{2 y}) * (\frac{- y}{x}) = - 1 .$

$\to 2 x = 1$ ________(5)

The point of intersection can be solve from eqn (1) and (2),

$y^{2} . y = k$

$\to y^{3} = k$

$\to y = k^{1 / 3}$

$∴ x = y^{2} = r^{2 / 3} .$

Hence, using eqⁿ (5) we get,

Upvote

Thumbs Down Icon

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..........

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

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:


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:

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

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:

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:

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 $\sqrt[]{3} r .$

Taking an Exam? Selecting a College?

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

On Shiksha, get access to

66K

Colleges

|

1.2K

Exams

|

6.9L

Reviews

|

1.8M

Answers

Learn more about...

Maths Ncert Solutions class 12th 2026

Maths Ncert Solutions class 12th 2026

View Exam Details

30 Answered Questions

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

or

See what others like you are asking & answering