If (x-a)²+(y-b)²=c², for some c>0, prove that [1+ (dy/dx )² ]³/2(d²y)/(dx²) is a constant independent of a and b.

MathsCBSE Class 12 · Sample paper, session 2024-25 · 5 marks · Long answer

CBSE Class 12 Mathematics sample paper 2024-25: question 34(b)

From the official CBSE sample question paper for 2024-25, section D, question 34(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34(b) · 5 marks · Long answerContinuity and Differentiability

If $(x - a)^{2} + (y - b)^{2} = c^{2}$ , for some $c > 0$ , prove that $\frac{[ 1 + ( \frac{d y}{d x} ) ^{2} ] ^{\frac{3}{2}}}{\frac{d ^{2} y}{d x ^{2}}}$ is a constant independent of $a$ and $b$ .

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

Open one hint at a time and try again before the next.

Hint 1

The relation is a circle with centre $(a, b)$ and radius $c$ . A parametric form makes the differentiation much easier.

Hint 2

Put $x - a = c cos θ$ and $y - b = c sin θ$ , then find $d y / d x$ in terms of $θ$ .

Hint 3

For $d^{2} y / d x^{2}$ , differentiate $d y / d x$ with respect to $θ$ and multiply by $d θ / d x$ . Then use $1 + cot^{2} θ = cosec^{2} θ$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

The relation is a circle with centre $(a, b)$ and radius $c$ , so take the parametric form $x - a = c cos θ$ and $y - b = c sin θ$ .
½ mark
Differentiate with respect to $θ$ : $\frac{d x}{d θ} = - c sin θ$ and $\frac{d y}{d θ} = c cos θ$ .
½ mark
Divide the two: $\frac{d y}{d x} = \frac{d y / d θ}{d x / d θ} = \frac{c cos θ}{- c sin θ} = - cot θ$ .
1 mark
Differentiate again with respect to $x$ using the chain rule: $\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d θ} (- cot θ) \cdot \frac{d θ}{d x} = cosec^{2} θ \cdot \frac{1}{- c sin θ}$ .
½ mark
Simplify: $\frac{d ^{2} y}{d x ^{2}} = - \frac{cosec ^{3} θ}{c}$ .
½ mark
Find the numerator: $[1 + (\frac{d y}{d x})^{2}]^{3/2} = (1 + cot^{2} θ)^{3/2} = (cosec^{2} θ)^{3/2} = cosec^{3} θ$ .
½ mark
Form the ratio: $\frac{[ 1 + ( \frac{d y}{d x} ) ^{2} ] ^{3/2}}{\frac{d ^{2} y}{d x ^{2}}} = \frac{cosec ^{3} θ}{- \frac{cosec ^{3} θ}{c}} = - c$ .
1 mark
The value $- c$ is a constant and contains neither $a$ nor $b$ . Hence the expression is a constant independent of $a$ and $b$ . Hence proved.
½ mark

Answer: The expression equals $- c$ , a constant independent of $a$ and $b$ .

Where the marks are

Uses the parametric substitution $x - a = c cos θ$ , $y - b = c sin θ$ (1/2 mark)
Finds $d x / d θ$ and $d y / d θ$ (1/2 mark)
Obtains $d y / d x = - cot θ$ (1 mark)
Differentiates again using the chain rule (1/2 mark)
Obtains $d^{2} y / d x^{2} = - cosec^{3} θ / c$ (1/2 mark)
Simplifies $(1 + cot^{2} θ)^{3/2}$ to $cosec^{3} θ$ (1/2 mark)
Gets the ratio $= - c$ (1 mark)
Concludes that the value is independent of $a$ and $b$ (1/2 mark)

Revise the chapter

Chapter: Continuity and Differentiability (notes, formulas, where marks are lost and every past question)
Topic: Parametric and Second Order Derivatives

Practise this chapterBoard-style questions from Continuity and Differentiability, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2024-25: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Mathematics sample paper 2024-25: question 34(b)

Hints

Step-by-step solution

Revise the chapter

More questions from Continuity and Differentiability