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Conic Sections Ncert Solutions Maths class 11th Maths Class 11th

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 b e \frac{5}{4} .$ If the equation of the normal at the point $(\frac{8}{\sqrt[]{5}}, \frac{12}{5})$ on the hyperbola is $8 \sqrt[]{5} x + β y = λ, t h e n λ - β$ is equal to……………..

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alok kumar singh | Contributor-Level 10

2 months ago

$e = \frac{5}{4}$

$b^{2} = a^{2} (e^{2} - 1)$

$\Rightarrow b = \frac{3 a}{4}$

$p (\frac{8}{\sqrt[]{5}}, \frac{12}{5})$

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

$\Rightarrow \frac{64}{5 a^{2}} - \frac{144}{25 b^{2}} = 1$

$\Rightarrow \frac{\sqrt[]{5} x}{3} + \frac{5}{8} y = \frac{8}{3} + \frac{3}{2} = \frac{25}{6}$

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Slope of axis = $\frac{1}{2}$

$y - 3 = \frac{1}{2} (x - 2)$

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⇒ 2y – x – 4 = 0

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Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 b e \frac{5}{4} .$ If the equation of the normal at the point $(\frac{8}{\sqrt[]{5}}, \frac{12}{5})$ on the hyperbola is $8 \sqrt[]{5} x + β y = λ, t h e n λ - β$ is equal to……………..

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Let the eccentricity of the hyperbola x 2 a 2 − y 2 b 2 = 1 b e 5 4 . If the equation of the normal at the point ( 8 5 , 1 2 5 ) on the hyperbola is 8 5 x + β y = λ , t h e n λ − β is equal to……………..

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Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 b e \frac{5}{4} .$ If the equation of the normal at the point $(\frac{8}{\sqrt[]{5}}, \frac{12}{5})$ on the hyperbola is $8 \sqrt[]{5} x + β y = λ, t h e n λ - β$ is equal to……………..