IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

VINEET_IIT_2009

FIND THE POINTS ON THE ELLIPSE x²/a²+y²/b²=1 SUCH THAT THE TANGENT AT EACH OF THEM MAKES EQUAL ANGLE WITH THE AXES.ALSO PROVE THAT THE LENGTH OF PERPENDICULARS FROM THE CENTRE TO THESE TANGENTS IS [(a²+b²)/2]^1/2.

Conjurer

Logically speaking , if the tangent makes equal angles with the axes it should have slope 1 or -1.

allamraju

There will be four points.They are ((+or-)a/rta²+b²,(+or-)b/rta²+b²).

Since the tangents make equal angles with axes,their slope must be (+ or -)1.

y=mx+c is tangent to ellipse iff c²=a²m²+b²=a²+b².So,the tangents are (+or-)x-y(+or-)rta²+b²=0.So,the length of the perpendicular from centre is rta²+b²/rt2

magiclko

the solution is on its way...approach is perfectly fyn....

do knck back in case u r unable to get the answer!!!

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There will be four points.They are ((+or-)a/rta2+b2,(+or-)b/rta2+b2).

Since the tangents make equal angles with axes,their slope must be (+ or -)1.

y=mx+c is tangent to ellipse iff c2=a2m2+b2=a2+b2.So,the tangents are (+or-)x-y(+or-)rta2+b2=0.So,the length of the perpendicular from centre is rta2+b2/rt2

There will be four points.They are ((+or-)a/rta²+b²,(+or-)b/rta²+b²).

y=mx+c is tangent to ellipse iff c²=a²m²+b²=a²+b².So,the tangents are (+or-)x-y(+or-)rta²+b²=0.So,the length of the perpendicular from centre is rta²+b²/rt2