ISI2015-MMA-46

Question

If the tangent at the point $P$ with coordinates $(h,k)$ on the curve $y^2=2x^3$ is perpendicular to the straight line $4x=3y$, then

• A. $(h,k) = (0,0)$
• B. $(h,k) = (1/8, -1/16)$
• C. $(h,k) = (0,0) \text{ or } (h,k) = (1/8, -1/16)$
• D. no such point $(h,k)$ exists

Answer 1

Important theorem: When two straight lines   $y_{1}=m_{1}x+c_{1}$  and   $y_{2}=m_{2}x+c_{2}$ are perpendicular to each other , the product of their respective slopes is $-$$1$  i.e.  $m_{1}\times m_{2}=-1$.

Now, for finding slope of the tangent at the given point $\left ( h,k \right )$,  we will find value of $\frac{dy}{dx}$ at the given point.

$\therefore$  $2$$y$$\frac{dy}{dx}$  $=$  $6$$x^{2}$   $\Rightarrow$   $\frac{dy}{dx}$  $=$   $\frac{3x^{2}}{y}$

The slope at the point $\left ( h,k \right )$  would be :   $\frac{3h^{2}}{k}$  ----$\left ( 1 \right )$

The slope of the given straight line is :    $\frac{4}{3}$      ----  $\left ( 2 \right )$

Using  $\left ( 1 \right )$  and   $\left ( 2 \right )$  and the above theorem,

$\frac{3h^{2}}{k}$  $\times$    $\frac{4}{3}$   $=$   $-$$1$    $\Rightarrow$  $k$  $=$  $-$$4$$h^{2}$  ----  $\left ( 3 \right )$

Also the point $\left ( h,k \right )$  satisfies the curve, so  $k^{2}=2h^{3}$  ---- $\left ( 4 \right )$

Putting the value of k from $\left ( 3 \right )$  into $\left ( 4 \right )$ , we get $16h^{4}=2h^{3}$

$\therefore$   $2h^{3}\left ( 8h-1 \right )=0$    $\Rightarrow$   $h=0$  or  $h=\frac{1}{8}$

So the respective value of k for h calculated above would be: $k=0$ or  $k=-\frac{1}{16}$

$\therefore$  Option C is the correct answer.