Question

A variable $\Delta ABC$ in the xy plane has its orthocentre at vertex 'B', a fixed vertex 'A' at the origin and the third vertex 'C' restricted to lie on the parabola $y=1+\frac{7x^2}{36}$ The point B starts at the point $(0,1)$ at time $t=0$ and moves upward along the y - axis at t a constant velocity of $2cm/sec$ How fast is the area of the triangle increasing when $t=\frac{7}{2}sec$

• A. $\frac{66}{7}$
• B. $\frac{86}{7}$
• C. $\frac{75}{7}$
• D. $\frac{89}{7}$

Answer 1

The correct option is A $\frac{66}{7}$

The parabola $y=1+\frac{7x^2}{36}$ intersects the $y$-axis at coordinates$\left(0,1\right)$. The third vertex C lies on the parabola, so, the coordinates of C will be $\left(x,1+\frac{7x^2}{36}\right)$ and the vertex A has its coordinates on the origin.

At $t=0$, $y=1$ and as the point B moves upward along the $y$-axis with constant speed of $2cm/sec$, so at $t=t$, $y=1+2t$. Therefore, the coordinates of B will be $\left(0,1+2t\right)$.

Now, $y_B=y_C$

$1+2t=1+\frac{7x^2}{36}$

$7x^2=2t\times 36$

$x^2=\frac{2t}{7}\times 36$

$x=\sqrt{\frac{2t}{7}}\times 6$

Join AC and BC where $BC=x$.

The area of $\Delta ABC$ is,

$A=\frac{1}{2}\times b\times h$

$=\frac{1}{2}\times x\times \left(1+2t\right)$

$=\frac{1}{2}\times \sqrt{\frac{2t}{7}}\times 6\times \left(1+2t\right)$

$=3\left(1+2t\right)\sqrt{\frac{2t}{7}}$

$\frac{dA}{dt}=3\left[\left(1+2t\right)\times \frac{2}{7}\times \frac{1}{2\sqrt{\frac{2t}{7}}}+\sqrt{\frac{2t}{7}}\left(2\right)\right]$

$=\frac{3}{7}\times \frac{1}{\sqrt{\frac{2t}{7}}}\times \left(1+2t\right)+6\sqrt{\frac{2t}{7}}$

At $t=\frac{7}{2}sec$,

$\frac{dA}{dt}=\frac{3}{7}\times \frac{1}{\sqrt{\frac{2}{7}\times \frac{7}{2}}}\times \left(1+2\times \frac{7}{2}\right)+6\sqrt{\frac{2}{7}\times \frac{7}{2}}$

$=\frac{3}{7}\times 8+6$

$=\frac{24+42}{7}$

$=\frac{66}{7}$