Maharashtra State BoardHSC Science (Electronics) 11th
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Find the equation of the tangent to the ellipse x25+y24 = 1 passing through the point (2, –2) - Mathematics and Statistics

Sum

Find the equation of the tangent to the ellipse `x^2/5 + y^2/4` = 1 passing through the point (2, –2)

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Solution

Given equation of the ellipse is `x^2/5 + y^2/4` = 1

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

a2 = 5 and b2 = 4

Equations of tangents to the ellipse

`x^2/"a"^2 + y^2/"b"^2` = 1 having slope m are

y = `"m"x ± sqrt("a"^2"m"^2 + "b"^2)`

Since (2, – 2) lies on both the tangents,

– 2 = `2"m" ± sqrt(5"m"^2 + 4)`

∴ – 2 –2m = `± sqrt(5"m"^2 + 4)`

Squaring both the sides, we get

4m2 + 8m + 4 = 5m2 + 4

∴ m2 – 8m = 0

∴ m(m – 8) = 0

∴ m = 0 or m = 8

These are the slopes of the required tangents.

∴ By slope point form y – y1 = m(x – x1), the equations of the tangents are

y + 2 = 0(x – 2) and y + 2 = 8(x – 2)

∴ y + 2 = 0 and y + 2 = 8x – 16

∴ y + 2 = 0 and 8x – y – 18 = 0.

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APPEARS IN

Balbharati Mathematics and Statistics 1 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 7 Conic Sections
Miscellaneous Exercise 7 | Q 2.17 | Page 178
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