Question

The triangle formed by the tangent to the parabola $y = x^{2}$at the point whose abscissa is $x_{0} = \left( x_{0} \in \lbrack 1,2\rbrack \right)$, the

$y - axis$ and the straight line $y = x_{0}^{2}$ has the greatest area if $x_{0} =$

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

2

Answer 2

Let $P\left( x_{0},x_{0}^{2} \right)$ be any point on the parabola

Equation of the tangent at $P\left( x_{0},x_{0}^{2} \right)$is $xx_{0} = \frac{1}{2}\left( y + x_{0}^{2} \right)$

⇒ $2xx_{0} - y - x_{0}^{2} = 0$

Tangent meet the y-axis at T $\left( 0, - x_{0}^{2} \right)$.

Hence the area of the triangle

$\Delta PTQ = \frac{1}{2}PQ \times QT$ = $\frac{1}{2} \times x_{0} \times 2x_{0}^{2}$

which increases in the interval [1,2] and hence is greatest when $x_{0} = 2$.