a circle is concentric with the eclipse. prove that the common tangent is inclined to the major axis at an angle x=tan^-1√r^2-b^2/a^2-r^2. hence find its length
Answers
Answer:
Step-by-step explanation:
The equation of ellipse be x²/a² + y²/b² = 1.
The equation of circle be x² + y² = r².
Equation of tangent at circle => xcosθ + ysinθ = r.
The line y = mx + c will be tangent to the eclipse x²/a² + y²/b² = 1 if c² = a²m² + b².
Hence the equation of tangent is y = mx+√a²m² + b²
If this is tangent to circle as well then perpendicular from (0, 0) is equal to r.
=> √a²m² + b² / √m²+1 = |r|
squaring on both sides
=> a²m² + b² = r²(m² + 1)
=> a²m² + b² = r²m² + r²
=> (a² - r²)m² = r² - b²
=> m = √[r² - b²/a² - r²]
=> Tanθ = √[r² - b²/a² - r²]
=> θ = Tan⁻¹(√[r² - b²/a² - r²])
Let P and Q be the points of contact of the common tangent with ellipse and circle respectively and O be the common centre of the two, then
PQ = √(OP² - OQ²) [∠CPQ = 90°]
The coordinates of P are [(-a²m)/√(a²m² + b²), b²/√(a²m² + b²)]
and co-ordinates of O are (0,0).
Length OP = √[(-a²m)/√(a²m² + b²)]² + [b²/√(a²m² + b²)]²
= √[a⁴m²/a²m² + b² + b⁴/a²m² + b²]
= √a⁴m² + b⁴/a²m² + b².
Length of OQ = r.
PQ = √(OP² - OQ²) = √ [a⁴m² + b⁴/a²m² + b² - r²]