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11th Math Practical No.7 Ellipse and hyperbola. Answer Solution | Maharashtra Board HSC Board | Balbharati

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11th Std Math Practical

Mathematics and Statistics (Arts and Science)

Answers (Solutions)

Practical  No.7

7) Ellipse and hyperbola

Practical Session No. 7-Ellipse and hyperbola



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11th-Math-Practical-No-7-Ellipse-and-hyperbola-answer-solution-Balbharati-Maharashtra-Board-HSC-Board
11th-Math-Practical-No-7-Ellipse-and-hyperbola-answer-solution-Balbharati-Maharashtra-Board-HSC-Board


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Practical Session No. 7
 Ellipse and hyperbola
1) Find the lengths of the major and minor axes, coordinates of vertices, eccentricity, co-ordinates of the foci, equations of directrices and the length of the Latus Rectum of the following conics
𝑖
) 9𝑥𝑥2+16𝑦𝑦2=144, 𝑖 ) 4𝑥𝑥2+25𝑦𝑦2=100
𝑖
) 𝑥𝑥225−𝑦𝑦29=1 𝑖 𝑖𝑖) 𝑦𝑦24−𝑥𝑥29=1
2) Find the equation of ellipse referred to its principal axes with eccentricity 34 and passing through the point (6,4).
3) An ellipse has OB as a semi-major axis, S and S’ are its foci and ∠𝑆𝑆𝑆 𝑆𝑆′ is a right- angle, then find the eccentricity of the ellipse.
4) Find focal distances of the point P􀵫5,4√3􀵯 on the ellipse 16𝑥𝑥2+25𝑦𝑦2=1600.
5) If 𝑒𝑒1 and 𝑒𝑒2 are eccentricities of hyperbolas 𝑥𝑥2𝑎𝑎2−𝑦𝑦2𝑏𝑏2=1 and 𝑦𝑦2𝑏𝑏2−𝑥𝑥2𝑎𝑎2=1
then show that 1𝑒𝑒12+1𝑒𝑒22=1
6) Find the equation of hyperbola whose
i) directrix is 2𝑥𝑥+𝑦𝑦=1 , focus at (1,2) and eccentricity √3.
ii) foci are at (±4 ,0) and the length of its latus rectum is 12 unit.
iii) vertices are ( 0 ,± 2 ) and the foci are at ( 0 ,± 3 ).
7) An interesting property of rectangular hyperbola
Equation of rectangular hyperbola is x y = k (k is non zero constant) …(I)
Tangent is drawn to the curve at point on it whose abscissa is ‘a’.
Therefore point of contact is (a, … )
Slope of tangent = 􁉂𝑑𝑑𝑑𝑑𝑑 􁉀𝑘𝑘𝑥𝑥􁉁􁉃 at point (a, … )
= 􁉀−𝑘𝑘𝑥𝑥2􁉁 at point (a, … )
= −𝑘𝑘𝑎𝑎2 …(II)
Equation of tangent to xy = k at point (a, … ) is (by slope point form)
y - ….. = 􁉀−𝑘𝑘𝑎𝑎2􁉁(x - a)
i.e. a2 (y - …) = -k (x – a)
i.e. …………
i.e. …………
this equation of tangent in terms of double intercept form is
𝑥𝑥2𝑎𝑎 + 𝑦𝑦2𝑘𝑘𝑎𝑎 = 1.
tangent cuts the x – axis at point P and the Y – axis at point Q.
clearly P = (… , 0) and Q = (0, …)
Area of Δ POQ = 12(OP)(OQ)
= 12(….)(….)
= 2k Verify this interesting result for different values of k.
i) xy = 4 at (2, 2) ii) xy = 12 at (-2, -6)

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