UPSEE 2017 Solved Paper

Given,
y=x2 ......(i)
x=

5

3

‌cos‌t,
y=

5

4

‌sin‌t .......(ii)
Which is parametric equation, we change this equation is cartesian equation as follows:
cos‌t=

3

5

x,sin‌t=

4

5

y
On squaring and adding both i.e. cos‌t and sin t, we get

9

25

x2+

16

25

y2=cos2t+sin2t
⇒9x2+16y2=25 ........(iii)
[∵cos2θ+sin2θ=1]
-The intersection points at Eq. (i) and (iii) are (1,1) and (−1,1).
Now, slope of tangent of Eq. (i) at point (1,1) is
m1=

dy

dx

=2x
∴ m1=

dy

dx

|(1,1)=2
And slope of tangent of Eq. (iii), at point (1,1) is
m2=

dy

dx

=−

9

16

∴ Angle at point of intersection of Eqs. (i) and (iii), we get
θ1=tan−1|

m1−m2

1+m1m2

|
θ1=tan−1|

2+ 9 16

1− 2×9 16

| =tan−1

41

2

Similarly, slope of tangent of Eq. (i) at point (−1,1)
m1=

dy

dx

|(−1,1)=−2
And slope of tangent of Eq. (iii) at point (−1,1)
m2=

dy

dx

=

−9

16

∴ Angle at point of intersection of Eqs. (i) and (iii), we get
θ2=tan−1|

−2−9 16

1− 18 16

| =tan−1

41

2