KEAM 2013 Math Paper

Given curve is ‌‌y2=px3+q....(i)
On differentiating w.r.t. x, we get
2y‌

dy

dx

=3px2‌‌⇒‌‌‌

dy

dx

=‌

3px2

2y

At (2,3),‌‌(‌

dy

dx

)(2,3)=‌

12p

6

=2p
since, line y=4x−5 is a tangent to the given curve.
∴ Slope =4=2p
⇒‌‌p=2
Now, put the value of p,x and y in Eq. (i), we get
(3)2=(2)(2)3+q
⇒‌‌9=16+q
⇒‌‌q=−7
Hence, ‌‌p+q=2−7=−5