KEAM 2015 Math Paper

Let the required point be P(x1,y1). The equation of the given curve is
y=3x2−11x−15
⇒‌‌‌

dy

dx

=6x−11
⇒(‌

dy

dx

)(x1,y1)=6x1−11
since, the tangent at P is parallel to the line joining (5,5) and (11,227) .
∴ Slope of the tangent at P= Slope of the line joining (5,5) and (11,227)
⇒‌‌(‌

dy

dx

)(x1,y1)=‌

227−5

11−5

⇒‌‌6x1−11=‌

222

6

⇒‌‌6x1−11=37
⇒6x1=48⇒x1=8