KEAM 2013 Math Paper

Given curves are
x=t2+3t−8
∴‌‌‌

dx

dt

=2t+3
and y=2t2−2t−5
∴‌‌‌

dy

dt

=4t−2
Slope of tangent =‌

dy

dx

=‌

dy

dt

×‌

dt

dx

=‌

4t−2

2t+3

......(i)
since, curve passes through the point (2,-1) .
∴‌‌t2+3t−8=2
and 2t2−2t−5=−1
⇒‌‌t2+3t−10=0
and 2t2−2t−4=0
⇒‌‌t2+5t−2t−10=0
and t2−t−2=0
⇒(t+5)(t−2)=0 and (t2−2t+t−2)=0
⇒t=−5,2 and (t−2)(t+1)=0
⇒‌‌t=−5,2 and t=−1,2
So, common value of t is 2 .
On putting t=2 in Eq. (i), we get
[‌

dy

dx

]‌at ‌t=2=‌

4(2)−2

2(2)+3

=‌

6

7

Slope of normal =‌

‌ −1 dy

dy dx

=‌

−1

(6∕7)

=−‌

7

6