SAURABH SIR HERE ALL THE DOUBTS :-​Q. THE CHORD OF THE PARABOLA-y = – a2x2 + 5ax – 4 TOUCHES THE CURVEy = 1/ 1-x AT THE POINT x=2 AND IS BISECTED BY THAT POINT . FIND “a”.Q . FOR THE CURVE x2/3 + y 2/3 = a2/3, SHOWTHAT | z |2 + 3p2 = a2WHEREz = x + iy AND pIS THE LENGTH OF THE PERPENDICULAR FROM (0,0) TO THE TANGENT AT ( x,y ) ON THE CURVE.Q . DETERMINE A DIFFERENTIABLE FUNCTIONy = f (x) WHICH SATISFIES f ‘ (x) = [ f (x) ]2 AND f(0) = – ½.​Q. IF THE TANGENT AT THE POINT (x1 , y1) TO THE CURVE x3 + y3 = a3 ( a not equal to 0) MEET THE CURVE AGAIN AT (x2,y2). THEN SHOW THATx2/ x1 + y2 / y1 = – 1.Q . SHOW THAT THE NORMALS TO THE CURVE -x = a ( cos t + t sin t ) ; y = a ( sin t – t cost ) ARE TANGENT LINES TO THE CIRCLE-x2+y2 =a2.

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