Sourabh Singh
Last Activity: 10 Years ago
Hii
For Q1)Q. THE CHORD OF THE PARABOLA-
y = – a2x2+ 5ax – 4 TOUCHES THE CURVE y = 1/ 1-x AT THE POINT x=2 AND IS BISECTED BY THAT POINT . FIND “a”.
Hint
You can use equation of chord in mid point form at the given point .And the chord is tangent to the curve at the given point . So find out slope of tangent at that point and equate it with the values of slope of chord which is in terms of “a“. Now you have 2 equations to find out the answers which you can easily find out.
Q2) FOR THE CURVE x2/3+ y2/3= a2/3, SHOW THAT | z |2+ 3p2= a2WHERE
z = x + iy AND p IS THE LENGTH OF THE PERPENDICULAR FROM (0,0) TO THE TANGENT AT ( x,y ) ON THE CURVE.
HINT
Use the subs. for x^(1/3) =m and y^(1/3) =n and you will end up with equation of circle whose tangent is easy to find out . And after that it’s just a question of slight modifications and you will get the answer.
Q3)
Q . DETERMINE A DIFFERENTIABLE FUNCTION y = f (x) WHICH SATISFIES
f ‘ (x) = [ f (x) ]2AND f(0) = – ½.
Hint
Take [ f (x) ]2 towards LHS and you can see that you will have(f ‘ (x) / [ f (x) ]2) which is very easy to integrate . Integrate both sides wrt x and you he condition given in the question to find out the specific answer.
I have given hints for above three . Take lessons from thses approach try to solve them again and still if unable to solve ask them again I will give hints for them also
Thanks