find the condition under which the lines x cos alpha + y Sin Alpha= p will be a tangent to the conic 3x^2+4y^2=5.
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in my keyboard there is no alpha sign so , I can use here ∅ . okay !!!!
3x² + 4y² = 5
xcos∅ + ysin∅ = p
x =( P- ysin∅)/cos∅
=Psec∅ -ytan∅
put this in curve
3(Psec∅ -ytan∅)² + 4y² = 5
3P²sec²∅ +3y²tan²∅ -6Pysec∅.tan∅ +4y² =5
( 4 + 3tan²∅)y² -6Pysec∅.tan∅ +3P²sec²∅ =0
(1+3sec²∅)y² -6Pysec∅.tan∅ +3P²sec²∅ =0
for tangents,
D = b² -4ac =0
(6Psec∅.tan∅)² -4(1+3sec²∅)(3P²sec²∅) =0
3P²tan²∅.sec²∅ -P²sec²∅ -3P²sec⁴∅ =0
3P²sec²∅(tan²∅ -sec²∅) -P²sec²∅=0
-4P²sec²∅ =0
Psec∅ =0
3x² + 4y² = 5
xcos∅ + ysin∅ = p
x =( P- ysin∅)/cos∅
=Psec∅ -ytan∅
put this in curve
3(Psec∅ -ytan∅)² + 4y² = 5
3P²sec²∅ +3y²tan²∅ -6Pysec∅.tan∅ +4y² =5
( 4 + 3tan²∅)y² -6Pysec∅.tan∅ +3P²sec²∅ =0
(1+3sec²∅)y² -6Pysec∅.tan∅ +3P²sec²∅ =0
for tangents,
D = b² -4ac =0
(6Psec∅.tan∅)² -4(1+3sec²∅)(3P²sec²∅) =0
3P²tan²∅.sec²∅ -P²sec²∅ -3P²sec⁴∅ =0
3P²sec²∅(tan²∅ -sec²∅) -P²sec²∅=0
-4P²sec²∅ =0
Psec∅ =0
vidya5:
ok and thank you very much
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