LE the common tangents to the curves <math> <mrow> <mn>4</mn> <mrow> <mo>(</mo> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>9</mn> </mrow> </math> and y2 = 4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and <math> <mrow> <mi mathvariant="script">l</mi> </mrow> </math> respectively denote the eccentricity and the length of the latus rectum of this ellipse, then <math> <mrow> <mfrac> <mrow> <mi mathvariant="script">l</mi> </mrow> <mrow> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math> is equal to……..

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Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Straight Lines Ncert Solutions Maths class 11th Maths Class 11th

LE the common tangents to the curves $4 (x^{2} + y^{2}) = 9$ and y² = 4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and $l$ respectively denote the eccentricity and the length of the latus rectum of this ellipse, then $\frac{l}{e^{2}}$ is equal to……..

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Raj Pandey | Contributor-Level 9

2 weeks ago

Let y = mx + c is the common tangent

$s o c = \frac{1}{m} = \pm \frac{3}{2} \sqrt[]{1 + m^{2}} \Rightarrow m^{2} = \frac{1}{3}$

so equation of common tangents will be

$y = \pm \frac{1}{\sqrt[]{3}} x \pm \sqrt[]{3}$

which intersects at Q (-3, 0)

Major axis and minor axis of ellipse are 12 and 6. So eccentricity

$e^{2} = 1 - \frac{1}{4} = \frac{3}{4}$

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Let y = mx + c is the common tangent

$s o c = \frac{1}{m} = \pm \frac{3}{2} \sqrt[]{1 + m^{2}} \Rightarrow m^{2} = \frac{1}{3}$

so equation of common tangents will be

$y = \pm \frac{1}{\sqrt[]{3}} x \pm \sqrt[]{3}$

which intersects at Q (-3, 0)

Major axis and minor axis of ellipse are 12 and 6. So eccentricity

$e^{2} = 1 - \frac{1}{4} = \frac{3}{4}$

and length of latus rectum

$= \frac{2 b^{2}}{a} = 3$

Hence $\frac{l}{e^{2}} = \frac{3}{3 / 4} = 4$

less

Let y = mx + c is the common tangent              <math> <mrow> <mi>s</mi> <mi>o</mi> <mtext> </mtext> <mtext> </mtext> <mi>c</mi> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mi>m</mi> </mrow> </mfrac> <mo>=</mo> <mo>±</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mroot> <mrow> <mn>1</mn> <mo>+</mo> <msup> <mrow> <mi>m</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow></mrow> </mroot> <mo>⇒</mo> <msup> <mrow> <mi>m</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math>                so equation of common tangents will be              <math> <mrow> <mi>y</mi> <mo>=</mo> <mo>±</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> <mi>x</mi> <mo>±</mo> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> </math>                which intersects at Q (-3, 0)              Major axis and minor axis of ellipse are 12 and 6. So eccentricity              <math> <mrow> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mn>1</mn> <mo>−</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> </mrow> </math>                and length of latus rectum              <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mi>a</mi> </mrow> </mfrac> <mo>=</mo> <mn>3</mn> </mrow> </math>                Hence <math> <mrow> <mfrac> <mrow> <mi mathvariant="script">l</mi> </mrow> <mrow> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>3</mn> <mo>/</mo> <mn>4</mn> </mrow> </mfrac> <mo>=</mo> <mn>4</mn> </mrow> </math>

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