is the slope of the tangent to the curve at P. The area of the triangle (in sq. units) formed by the tangent and the normal to the curve at P with X-axis is
(x−5) ⇒‌2y−4‌=7x−35 ⇒‌7x−2y‌=31 ∴ The tangent intersect the X-axis, y‌=0 ∴‌‌7x‌=31⇒x=‌
31
7
∴ Tangent intersect the X-axis at A(‌
31
7
,0) Now, slope of normal at P is −‌
2
7
Equation of normal at P is y−2=−‌
2
7
(x−5) ⇒‌‌7y−14=−2x+10⇒2x+7y=24 ∴x-intercept of normal, y=0 ⇒‌‌x=12 ⇒ The normal intersects the X-axis at B(15,0) Base of the triangle =|(12−‌