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If (x/a) 2 +(y/b) 2 =1(a>b) and x 2 -y 2 =c 2 cut at right angles, then a. a 2 +b 2 =2c 2 b. b 2 -a 2 =2c 2 c.a 2 -b 2 =2c 2 d. a 2 -b 2 =2c 2

If (x/a)2 +(y/b)2=1(a>b) and x2 -y2 =c2 cut at right angles, then


a. a2+b2=2c2


b. b2-a2=2c2


c.a2 -b2=2c2


d. a2 -b2=2c2

Grade:12

2 Answers

Badiuddin askIITians.ismu Expert
148 Points
12 years ago
Dear Vaibhav let a point on ellipse is (acosT ,bsinT)(let this is point where curve intersect each other) so normal at this point on ellipse,and tangent on hyperbola will be same line. normal on ellipse axsecT - by cosecT =a^2 -b^2 ..............1 tangent on parabola at (acosT ,bsinT) axcosT - by sinT =c^2 .....................2 equation 1 and 2 represent the same equation so asecT /acosT = bcosecT /b sinT = (a^2 -b^2) /c^2 so asecT /acosT = bcosecT /b sinT cos^2T = sin^2T or tanT =1 or T =45 and bcosecT /b sinT = (a^2 -b^2) /c^2 a2 -b2=2c^2 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation. All the best. Regards, Askiitians Experts Badiuddin
Badiuddin askIITians.ismu Expert
148 Points
12 years ago

Dear Vaibhav l

et a point on ellipse is (acosΘ ,bsinΘ)(let this is point where curve intersect each other)

 so normal at this point on ellipse,and tangent on hyperbola will be same line.

 normal on ellipse axsecΘ - by cosecΘ =a^2 -b^2 ..............1

tangent on parabola at (acosΘ ,bsinΘ) is  axcosΘ - by sinΘ =c^2 .....................2

equation 1 and 2 represent the same equation

so asecΘ /acosΘ= bcosecΘ /b sinΘ = (a^2 -b^2) /c^2

so asecΘ /acosΘ= bcosecΘ /b sinΘ

 cos^2Θ= sin^2Θ

or tanΘ =1

 or Θ =45

and bcosecΘ /b sinΘ = (a^2 -b^2) /c^2

a2 -b2=2c^2

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation.

All the best. Regards,

Askiitians Experts

 Badiuddin

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