KEAM 2014 Math Paper

Given equation of curve is y2=4x+5
On differentiating both sides w.r.t. x, we get
2y‌

dy

dx

=4
At point (x1,y1)
‌

dy

dx

=‌

2

y1

=m1  (say)
Also, given straight line is 2x−y+5=0
So, the slope of this line is m2=2
since, the line is parallel to the tangent of the curve.
∴‌‌m1=m2⇒‌

2

Y1

=2⇒y1=1
As the point (x1,y1) lies on the given curve.
∴‌‌y12=4x1+5
Put y1=1, we get
12=4x1+5
⇒‌‌4x1=−4
⇒‌‌x1=−1
Hence, the point of contact is (-1,1)