Find minimum distance between the circle x² + (y-3)² = 1 and parabola y² = 4x

Equation of normal <!-- [if gte msEquation 12]>y=mx-2m-m3

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Maths NCERT Exemplar Solutions Class 11th Chapter Two Conic Sections Ncert Solutions Maths class 11th Maths Class 11th

Find minimum distance between the circle x² + (y-3)² = 1 and parabola y² = 4x

Option 1 -

√2

Option 2 -

Option 3 -

√2 - 1

Option 4 -

2 + 1

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1 Answer

Answered by

Raj Pandey | Contributor-Level 9

a month ago

Correct Option - 3

Detailed Solution:

Equation of normal $y = m x - 2 m - m^{3}$ must pass through centre $(0,3)$

$\begin{array}{r} m^{3} + 2 m + 3 = 0 \\ m^{2} (m + 1) - m (m + 1) + 3 \cdot 1 (m + 1) = 0 \\ (m + 1) (m^{2} - m + 3) = 0 \\ m = - 1 \end{array}$

$∴$ point of contact of normal at parabola is $({a m}^{2}, - 2 a m) = (1,2)$

So distance between parabola and circle is $= \sqrt[]{2} - 1$

Equation of normal    <math> <mi>y</mi> <mo>=</mo> <mi>m</mi> <mi>x</mi> <mo>-</mo> <mn>2</mn> <mi>m</mi> <mo>-</mo> <msup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math> must pass through centre <math> <mo>(</mo> <mn>0,3</mn> <mo>)</mo> </math>   <math> <mtable columnalign="right"> <mtr> <mtd> <mrow></mrow> </mtd> <mtd> <mrow> <msup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> <mo>+</mo> <mn>2</mn> <mi>m</mi> <mo>+</mo> <mn>3</mn> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> <mtd> <mrow> <msup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>(</mo> <mi>m</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> <mo>-</mo> <mi>m</mi> <mo>(</mo> <mi>m</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> <mo>+</mo> <mn>3</mn> <mo>⋅</mo> <mn>1</mn> <mo>(</mo> <mi>m</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> <mtd> <mrow> <mi> </mi> <mo>(</mo> <mi>m</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> <mfenced separators="|"> <mrow> <mrow> <msup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>-</mo> <mi>m</mi> <mo>+</mo> <mn>3</mn> </mrow> </mrow> </mfenced> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> <mtd> <mrow> <mi>m</mi> <mo>=</mo> <mo>-</mo> <mn>1</mn> </mrow> </mtd> </mtr> </mtable> </math>    <math> <mo>∴</mo> <mrow> <mi mathvariant="normal"> </mi> </mrow> </math>  point of contact of normal at parabola is <math> <mfenced separators="|"> <mrow> <mrow> <msup> <mrow> <mrow> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>,</mo> <mo>-</mo> <mn>2</mn> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">m</mi> </mrow> </mrow> </mfenced> <mo>=</mo> <mo>(</mo> <mn>1,2</mn> <mo>)</mo> </math>  <div><div><picture><source srcset="" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1757574389php3Fgmlo.jpeg" alt="" width="141" height="106" loading="lazy"></picture></div></div>So distance between parabola and circle is <math> <mo>=</mo> <mroot> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow></mrow> </mroot> <mo>-</mo> <mn>1</mn> </math>

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⇒ 2y – 6 = x – 2

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