Let <math><mi>P</mi><mo>(</mo><mn>3,3</mn><mo>)</mo></math> be a point on the hyperbola, <math><mfrac><mrow><mrow><msup><mrow><mrow><mi>x</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><msup><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac><mo>-</mo><mfrac><mrow><mrow><msup><mrow><mrow><mi>y</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><msup><mrow><mrow><mi>b</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac><mo>=</mo><mn>1</mn></math> . If the normal to it at <math><mi>P</mi></math> intersects the <math><mi>x</mi></math> -axis at <math><mo>(</mo><mn>9,0</mn><mo>)</mo></math> and <math><mi>e</mi></math> is its eccentricity, then the ordered pair <math><mfenced separators="|"><mrow><mrow><msup><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>,</mo><msup><mrow><mrow><mi>e</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfenced></math> is equal to:

Since (3,3) lies on x2a2-y2b2=19a2-9b2=1Now, normal at (3,3) is y-3=-a2b2 (x-3) , which passes through (9,0)⇒b2=2a2So, e2=1+b2a2=3Also, a2=92 (From (i) and (ii)Thus, a2, e2=92, 3

Let&nbsp;<math><mi>P</mi><mo>(</mo><mn>3,3</mn><mo>)</mo></math>&nbsp;be a point on the hyperbola,&nbsp;<math><mfrac><mrow><mrow><msup><mrow><mrow><mi>x</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><msup><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac><mo>-</mo><mfrac><mrow><mrow><msup><mrow><mrow><mi>y</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><msup><mrow><mrow><mi>b</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac><mo>=</mo><mn>1</mn></math>&nbsp;. If the normal to it at&nbsp;<math><mi>P</mi></math>&nbsp;intersects the&nbsp;<math><mi>x</mi></math>&nbsp;-axis at&nbsp;<math><mo>(</mo><mn>9,0</mn><mo>)</mo></math>&nbsp;and&nbsp;<math><mi>e</mi></math>&nbsp;is its eccentricity, then the ordered pair&nbsp;<math><mfenced separators="|"><mrow><mrow><msup><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>,</mo><msup><mrow><mrow><mi>e</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfenced></math>&nbsp;is equal to:

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Maths Class 12th Conic Sections Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let $P (3,3)$ be a point on the hyperbola, $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ . If the normal to it at $P$ intersects the $x$ -axis at $(9,0)$ and $e$ is its eccentricity, then the ordered pair $(a^{2}, e^{2})$ is equal to:

Option 1 -

$(\frac{3}{2}, 2)$

Option 2 -

$(\frac{9}{2}, 2)$

Option 3 -

$(9,3)$

Option 4 -

$(\frac{9}{2}, 3)$

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

Since $(3,3)$ lies on $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$\frac{9}{a^{2}} - \frac{9}{b^{2}} = 1$

Now, normal at $(3,3)$ is $y - 3 = - \frac{a^{2}}{b^{2}} (x - 3)$ ,

which passes through $(9,0) \Rightarrow b^{2} = 2 a^{2}$

So, $e^{2} = 1 + \frac{b^{2}}{a^{2}} = 3$

Also, $a^{2} = \frac{9}{2}$

(From (i) and (ii)

Thus, $(a^{2}, e^{2}) = (\frac{9}{2}, 3)$

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Let $P (3,3)$ be a point on the hyperbola, $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ . If the normal to it at $P$ intersects the $x$ -axis at $(9,0)$ and $e$ is its eccentricity, then the ordered pair $(a^{2}, e^{2})$ is equal to:

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Let P(3,3) be a point on the hyperbola, x2a2-y2b2=1 . If the normal to it at P intersects the x -axis at (9,0) and e is its eccentricity, then the ordered pair a2,e2 is equal to:

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Let $P (3,3)$ be a point on the hyperbola, $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ . If the normal to it at $P$ intersects the $x$ -axis at $(9,0)$ and $e$ is its eccentricity, then the ordered pair $(a^{2}, e^{2})$ is equal to: