हिंदी

Find the equations of the tangents to the ellipse x216+y29 = 1, making equal intercepts on co-ordinate axes

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प्रश्न

Find the equations of the tangents to the ellipse `x^2/16 + y^2/9` = 1, making equal intercepts on co-ordinate axes

योग
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उत्तर

The equation of the ellipse is `x^2/16 + y^2/9` = 1

Comparing with `x^2/"a"^2 + y^2/"b"^2` = 1, we get,

a2 = 16, b2 = 9

Let the required tangent make intercept c on each axis

∴ its equation is `x/"c" + y/"c"` = 1

∴ x + y = c     ...(I)

∴ y = – x + c

We know that y = mx + c is a tangent to the ellipse

`x^2/"a"^2 + y^2/"b"^2` = 1 if c2 = a2m2 + b

Here, a2 = 16, b2 = 9, m = – 1

∴ c2 = 16 × 1 + 9 = 25

∴ c = ± 5

∴ Using (I), the required equations of tangents are x + y = ± 5.

Alternate Method:

The equation of the ellipse is `x^2/16 + y^2/9` = 1

Comparing this with `x^2/"a"^2 + y^2/"b"^2` = 1, we get,

a2 = 16, b2 = 9

The equation of tangents to the ellipse `x^2/"a"^2 + y^2/"b"^2` = 1 with slope m are

y = `"m"x  ± sqrt("a"^2"m"^2 + "b"^2)`

The tangent makes equal intercepts on the coordinate axes, its slope = m = – 1

∴ `sqrt("a"^2"m"^2 + "b"^2) = sqrt(16(-1)^2 + 9)` = 5

∴ the equations of required tangents are y = –x ± 5

∴ x + y = ± 5

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अध्याय 7: Conic Sections - Exercise 7.2 [पृष्ठ १६४]

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बालभारती Mathematics and Statistics (Arts and Science) Part 1 [English] Standard 11 Maharashtra State Board
अध्याय 7 Conic Sections
Exercise 7.2 | Q 17 | पृष्ठ १६४

संबंधित प्रश्न

Find the

  1. lengths of the principal axes.
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  4. length of the latus rectum
  5. distance between focii
  6. distance between directrices of the ellipse:

3x2 + 4y2 = 12


Find the 

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