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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

If p1 and p2 are the lengths of the perpendiculars from the origin to the straight lines x sec θ + y cosec θ = 2a and x cos θ – y sin θ = a cos 2θ, then prove that p12 + p22 = a2

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प्रश्न

If p1 and p2 are the lengths of the perpendiculars from the origin to the straight lines x sec θ + y cosec θ = 2a and x cos θ – y sin θ = a cos 2θ, then prove that p12 + p22 = a

बेरीज
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उत्तर

Given P1 is the length of the perpendicular from the origin to the straight line

x sec θ + y cosec θ – 2a = 0

P1 = `(0 * sec theta + 0 * "cosec"  theta - 2"a")/sqrt(sec^2theta + "cosec"^2theta)`

P12 = `(4"a"^2)/(sec^2theta + "cosec"^2theta)`  .....(1)

Also given P2 is the length of the perpendicular from the origin to the straight line

x cos θ – y sin θ – a cos 2θ = 0

P2 = `(0 * cos theta - 0 sin theta - "a" cos 2theta)/sqrt(cos^2theta + (- sin theta)^2`

P2 = `(- "a" cos 2theta)/sqrt(cos^2theta + sin^2theta)`

= – a cos 2θ

P22 = a2 cos2 2θ  ......(2)

P12 + P22 = `(4"a"^2)/(sec^2theta + "cosec"^2theta) + "a"^2cos^2 2theta`

= `(4"a"^2)/(1/(cos^2theta) + 1/(sin^2theta)) + "a"^2(cos^2theta - sin^2theta)`

= `(4"a"^2)/((sin^2theta + cos^2theta)/(cos^2theta * sin^2theta)) + "a"^2(cos^4theta + sin^4theta - 2cos^2theta sin^2theta)`

= 4a2 cos2θ sin2θ + a2cos4θ + a2sin4θ - 2a2cos2θ sin2θ

= a2cos4θ + a2sin4θ + 2a2 cos2θ sin2θ

= a2[cos4θ + sin4θ + 2cos2θ sin2θ]

= a2[cos2θ + sin2θ]2

P12 + P22 = a2 

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Angle Between Two Straight Lines
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Two Dimensional Analytical Geometry - Exercise 6.3 [पृष्ठ २७२]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 6 Two Dimensional Analytical Geometry
Exercise 6.3 | Q 11 | पृष्ठ २७२

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