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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
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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 + b2
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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संबंधित प्रश्न
Answer the following:
Find the
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- co-ordinates of the foci
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